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Commit 53adae2

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新增二分题目
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‎README.md‎

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@@ -32,17 +32,21 @@
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| 36 | [有效的数独](https://leetcode-cn.com/problems/valid-sudoku/)| [isValidSudoku](./other/leetcode/medium/isValidSudoku.h) | <font color=orange> medium </font> | ✅|
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| 39 | [组合总和](https://leetcode-cn.com/problems/combination-sum/)| [combinationSum](./backtracking/leetcode/medium/combinationSum.h) | <font color=orange> medium </font> | ✅|
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| 46 | [全排列](https://leetcode-cn.com/problems/permutations/)| [permutations](./backtracking/leetcode/medium/permutations.h) | <font color=orange> medium </font> | ✅|
35+
| 50 | [Pow(x, n)](https://leetcode-cn.com/problems/powx-n/solution/powx-n-by-leetcode/) | [powx](./bsearch/leetcode/medium/powx.h) | <font color=orange> medium </font> | ✅ |
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| 53 | [最大子序和](https://leetcode-cn.com/problems/maximum-subarray/) | [maxSubArray](./array/leetcode/easy/maxSubArray.h) | <font color=green>easy</font> | ✅ |
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| 56 | [合并区间](https://leetcode-cn.com/problems/merge-intervals/) | [merge_intervals](./sort/leetcode/merge_intervals.h) | <font color=orange> medium </font> | ✅ |
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| 58| [最后一个单词的长度](https://leetcode-cn.com/problems/length-of-last-word) | [lengthOfLastWord](./string/leetcode/easy/lengthOfLastWord.h) | <font color=green>easy</font> | ✅ |
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| 66 | [加一](https://leetcode-cn.com/problems/plus-one/) | [plusOne](./array/leetcode/easy/plusOne.h) | <font color=green>easy</font> | ✅ |
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| 67 | [二进制求和](https://leetcode-cn.com/problems/add-binary/) | [addBinary](./array/leetcode/easy/addBinary.h) | <font color=green>easy</font> | ✅ |
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| 69 | [x 的平方根](https://leetcode-cn.com/problems/sqrtx/%E2%80%A8)| [mySqrt](./bsearch/leetcode/mySqrt.h) | <font color=green>easy</font> | ✅ |
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| 70 | [爬楼梯](https://leetcode-cn.com/problems/climbing-stairs/)| [climbStairs](./dp/leetcode/easy/climbStairs.h) | <font color=green>easy</font> | ✅ |
43+
| 74 | [搜索二维矩阵](https://leetcode-cn.com/problems/search-a-2d-matrix/) | [search_a_2d_matrix](./bsearch/leetcode/medium/search_a_2d_matrix.h) | <font color=orange> medium </font> | ✅ |
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| 75 | [颜色分类](https://leetcode-cn.com/problems/sort-colors/) | [sort_colors](./sort/leetcode/sort_colors.h) | <font color=orange> medium </font> | ✅ |
45+
| 81 | [搜索旋转排序数组II](https://leetcode-cn.com/problems/search-in-rotated-sorted-array-ii/) | [searchInRotatedSortedArrayII](./bsearch/leetcode/medium/searchInRotatedSortedArrayII.h) | <font color=orange> medium </font> | ✅ |
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| 83 | [删除排序链表中的重复元素](https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/)| [deleteDuplicates](./linkedList/leetcode/easy/deleteDuplicates.h) | <font color=green>easy</font> | ✅ |
4447
| 88 | [合并两个有序数组](https://leetcode-cn.com/problems/merge-sorted-array/) | [merge](./array/leetcode/easy/merge.h) | <font color=green>easy</font> | ✅ |
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| 100 | [相同的树](https://leetcode-cn.com/problems/same-tree/) | [isSameTree](./tree/leetcode/easy/isSameTree.h) | <font color=green>easy</font> | ✅ |
49+
| 153 | [寻找旋转排序数组中的最小值](https://leetcode-cn.com/problems/find-minimum-in-rotated-sorted-array/) | [find_minimum_in_rotated_sorted_array](./bsearch/leetcode/medium/find_minimum_in_rotated_sorted_array.h) | <font color=orange> medium </font> | ✅ |
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| 687 | [最长同值路径](https://leetcode-cn.com/problems/longest-univalue-path/)| [longestUnivaluePath](./recursion/leetcode/medium/longestUnivaluePath.h) | <font color=green>easy</font> | ✅ |
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# **算法思想**
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## [递归](./recursion.md) 🚶🚶🚶🚶
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### 什么是递归?
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试想一下电影院自己在第几排的场景, 想要知道我在第几排,那么就要去问前面的人他处于第几排,那么前面的人又怎么知道他自己在第几排呢?那自然是他也要去问他前面的人咯.
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这就是递归的场景,去的过程叫做<font size=5 color=red>"递"</font>, 回来的过程叫做<font size=5 color=red>"归"</font>,有来有回。
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![](./res/recursion_example.png)
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<font size=5 color=red>递归其实就是利用栈的数据结构,再加上一些简单的逻辑算法从而完成了问题的求解。只不过这个栈是由系统来提供的,我们只是无感知罢了.</font>
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### 适合场景
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满足下面两点:
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* 一个问题的解可以分解为几个子问题(规模更小的问题)的解, 并且子问题和问题除了数据规模不一样,求解思路是完全一样的。
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* 存在终止条件
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那么如何编写递归代码呢?
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关键在于找到将大问题分解为小问题的规律,写出递归公式,然后在推敲出终止条件。
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递归代码简洁高效,但是容易出现堆栈溢出,重复计算,函数调用耗时多,空间复杂度高等问题。一般嵌套比较少的场景可以使用递归。
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### [递归题目](./recursion.md)
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## [递归](./recursion.md)
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## [排序](./sort.md)
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## [二分查找](./bsearch.md)
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## [二分查找](./bsearch.md) 🚶🚶🚶🚶
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# 解决多阶段决策最优解模型的算法
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<!--# 解决多阶段决策最优解模型的算法
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解决问题的过程中,需要经过多个决策阶段,每个决策都会对应一个状态。我们寻找一组决策序列,经过这组决策序列,能够产生最终期望的最优值。我们把这种问题模型称为<font size=5 color=red>多阶段决策最优解模型</font>. DP,回溯,贪心都可以解决这类问题.
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利用动态规划解决的问题,需要满足三个特征:
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所以贪心算法能否解决算法问题的关键在于: 局部最优能不能达到全局最优?
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回溯算法是"万金油"。基本上贪心和dp能解决的问题,回溯都能解决。回溯相当于穷举搜索,列举出所有的情况,然后对比得到最优解。不过回溯的复杂度一般都是指数级的,只能用来解决小规模数据的问题。
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回溯算法是"万金油"。基本上贪心和dp能解决的问题,回溯都能解决。回溯相当于穷举搜索,列举出所有的情况,然后对比得到最优解。不过回溯的复杂度一般都是指数级的,只能用来解决小规模数据的问题。-->
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## [动态规划](./dp.md)

‎alg-cpp.xcodeproj/project.pbxproj‎

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096E161A2321681C00444948 /* search_a_2d_matrix.h */ = {isa = PBXFileReference; lastKnownFileType = sourcecode.c.h; path = search_a_2d_matrix.h; sourceTree = "<group>"; };
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3A5C8BDD22E1CEC400354740 /* searchInRotatedSortedArray.h */,
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095F60122322B3390072CF0C /* searchInRotatedSortedArrayII.h */,
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098F84CC23234824001219CB /* find_minimum_in_rotated_sorted_array.h */,
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09421FFA23142AEB00A7BA67 /* searchRange.h */,
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096E161723215E7300444948 /* powx.h */,
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096E161A2321681C00444948 /* search_a_2d_matrix.h */,
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path = medium;
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sourceTree = "<group>";
5.94 KB
Binary file not shown.

‎alg-cpp.xcodeproj/xcuserdata/junlongj.xcuserdatad/xcschemes/xcschememanagement.plist‎

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<key>base.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>7</integer>
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<integer>5</integer>
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</dict>
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<key>bsearch.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>8</integer>
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<integer>9</integer>
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</dict>
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<key>divideandconquer.xcscheme_^#shared#^_</key>
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<dict>
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<key>dp.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>12</integer>
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<integer>10</integer>
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</dict>
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<key>hasTable.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>10</integer>
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<integer>11</integer>
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</dict>
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<key>linkedList.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>9</integer>
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<integer>6</integer>
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</dict>
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<key>other.xcscheme_^#shared#^_</key>
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<dict>
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<key>sort.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>5</integer>
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<integer>8</integer>
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</dict>
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<key>stack+queue.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>6</integer>
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<integer>7</integer>
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</dict>
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<key>string.xcscheme_^#shared#^_</key>
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<dict>
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<key>tree.xcscheme_^#shared#^_</key>
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<dict>
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<key>orderHint</key>
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<integer>11</integer>
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<integer>12</integer>
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</dict>
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</dict>
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<key>SuppressBuildableAutocreation</key>

‎bsearch.md‎

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| &emsp;题号&emsp; | 题目链接&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;| 答案链接&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;&emsp;| &emsp;难度&emsp; | &emsp;完成度&emsp; |
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| :--: | :--: | :----------------------------------------------------------- | :----------------------------------------------------------- | :------: |
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| 33 | [搜索旋转排序数组](https://leetcode-cn.com/problems/search-in-rotated-sorted-array)| [searchInRotatedSortedArray](./bsearch/leetcode/medium/searchInRotatedSortedArray.h) | ✨✨ | ✅|
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| 50 | [Pow(x, n)](https://leetcode-cn.com/problems/powx-n/solution/powx-n-by-leetcode/) | [powx](./bsearch/leetcode/medium/powx.h) | <font color=orange> medium </font> | ✅ |
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| 69 | [x 的平方根](https://leetcode-cn.com/problems/sqrtx/%E2%80%A8)| [mySqrt](./bsearch/leetcode/mySqrt.h) | ✨ | ✅ |
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| 74 | [搜索二维矩阵](https://leetcode-cn.com/problems/search-a-2d-matrix/) | [search_a_2d_matrix](./bsearch/leetcode/medium/search_a_2d_matrix.h) | <font color=orange> medium </font> | ✅ |
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| 81 | [搜索旋转排序数组II](https://leetcode-cn.com/problems/search-in-rotated-sorted-array-ii/) | [searchInRotatedSortedArrayII](./bsearch/leetcode/medium/searchInRotatedSortedArrayII.h) | <font color=orange> medium </font> | ✅ |
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| 153 | [寻找旋转排序数组中的最小值](https://leetcode-cn.com/problems/find-minimum-in-rotated-sorted-array/) | [find_minimum_in_rotated_sorted_array](./bsearch/leetcode/medium/find_minimum_in_rotated_sorted_array.h) | <font color=orange> medium </font> | ✅ |
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| 367 | [有效的完全平方数](https://leetcode-cn.com/problems/valid-perfect-square/)| [isPerfectSquare](./bsearch/leetcode/isPerfectSquare.h) | ✨ | ✅|
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//
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// find_minimum_in_rotated_sorted_array.h
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// bsearch
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//
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// Created by junl on 2019年9月7日.
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// Copyright © 2019 junl. All rights reserved.
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//
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#ifndef find_minimum_in_rotated_sorted_array_hpp
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#define find_minimum_in_rotated_sorted_array_hpp
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#include <stdio.h>
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/*
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153.寻找旋转排序数组中的最小值
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假设按照升序排序的数组在预先未知的某个点上进行了旋转。
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( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
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请找出其中最小的元素。
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你可以假设数组中不存在重复元素。
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示例 1:
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输入: [3,4,5,1,2]
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输出: 1
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示例 2:
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输入: [4,5,6,7,0,1,2]
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输出: 0
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来源:力扣(LeetCode)
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链接:https://leetcode-cn.com/problems/find-minimum-in-rotated-sorted-array
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著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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*/
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namespace leetcode {
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using namespace std;
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/*
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思路:
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寻找波谷,即寻找数组相邻两个元素,前一个元素比后一个元素大的位置.
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1.我们首先判断[left....right]这部分是否已经是有序数组了,如果是有序数组,那么直接返回左边的数。
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2.如果不是有序的,那么久存在旋转的部分,最小数字肯定出现在旋转部分。
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note:注意寻找波谷的条件nums[mid-1] > nums[mid] || nums[mid] > nums[mid+1]
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*/
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int findMin(vector<int>& nums) {
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if (nums.empty()) return -1;
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if (nums.size()==1) return nums[0];
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int left=0;
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int right=nums.size()-1;
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while(left<=right){
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int mid = left+((right-left)>>1);
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//如果已经是有序部分,返回左节点
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if (nums[left] <= nums[mid] && nums[mid] <= nums[right]){
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return nums[left];
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}
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//寻找波谷
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if (mid >0 && nums[mid-1] > nums[mid]){
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return nums[mid];
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}
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if (mid +1 <nums.size() && nums[mid] > nums[mid+1]) {
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return nums[mid+1];
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}
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//查找哪半边是有序的
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if (nums[mid] > nums[left]){
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//左边是有序的
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left = mid + 1;
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}else{
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//右边是有序的
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right = mid - 1;
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}
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}
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return -1;
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}
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}
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#endif /* find_minimum_in_rotated_sorted_array_hpp */

‎bsearch/leetcode/medium/powx.h‎

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//
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// powx.h
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// bsearch
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//
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// Created by junl on 2019年9月5日.
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// Copyright © 2019 junl. All rights reserved.
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//
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#ifndef powx_hpp
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#define powx_hpp
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#include <stdio.h>
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/*
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实现 pow(x, n) ,即计算 x 的 n 次幂函数。
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示例 1:
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输入: 2.00000, 10
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输出: 1024.00000
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示例 2:
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输入: 2.10000, 3
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输出: 9.26100
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示例 3:
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输入: 2.00000, -2
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输出: 0.25000
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解释: 2-2 = 1/22 = 1/4 = 0.25
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说明:
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-100.0 < x < 100.0
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n 是 32 位有符号整数,其数值范围是 [−231, 231 − 1] 。
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来源:力扣(LeetCode)
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链接:https://leetcode-cn.com/problems/powx-n
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著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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*/
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namespace leetcode {
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/*
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思路:
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常规思路,O(n)循环,每次乘以x,当n很大时,会超时.
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*/
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//bad
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double myPow2(double x, int n){
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long N = n; //这里用long来保存,因为-n可能会导致数据溢出。
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if (N < 0) {
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x = 1 / x;
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N = -N;
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}
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double ans = 1;
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for (long i = 0; i < N; i++)
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ans = ans * x;
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return ans;
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}
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/*
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思路二:
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利用x^n * x^n = x^(2n)这个公式。如果我们已知x^(n/2)的话,那么我们可以直接计算出x^n了。那怎么求解x^(n/2)呢,又通过x^(n/4),这不就是递归吗?
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*/
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#pragma mark - 快速幂递归
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//good
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double fastPow(double x , long n){
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if (n == 0) {
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return 1.0;
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}
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double r = fastPow(x, n/ 2);
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if (n % 2 == 0) {
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//如果是偶数的话
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return r * r;
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}else{
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return r * r * x;//比如5 = 2 * 2 + 1
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}
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}
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double myPow(double x, int n){
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long N = n; //这里用long来保存,因为-n可能会导致数据溢出。
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if (N < 0) {
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x = 1 / x;
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N = -N;
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}
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return fastPow(x, N);
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}
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#pragma mark - 快速幂循环
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/*
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思路:
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现在我们要计算x^n。我们把n看成二进制编码,那么N可以利用二进制运算计算得出。比如:
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x = 2 , n = 5, 0101
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那么 x^n = x^1 * x^4,即当二进制当前位i的值为1时,我们需要乘以x^(2^i).
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*/
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double myPow3(double x, int n){
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long N = n; //这里用long来保存,因为-n可能会导致数据溢出。
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if (N < 0) {
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x = 1 / x;
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N = -N;
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}
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double result = 1.0;
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double ctPositionValue = x;
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for (long i=n; i>0; i>>=1) {
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if (i & 1 ) {//如果末尾为1的话
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result *= ctPositionValue;
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}
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ctPositionValue*=ctPositionValue;//1,2,4,8,16
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}
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return result;
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}
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void test_pow(){
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std::cout << myPow3(2.0,INT32_MAX) << std::endl;
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std::cout << myPow3(2.0,20) << std::endl;
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std::cout << myPow2(2.0,20) << std::endl;
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}
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}
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#endif /* powx_hpp */

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