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Commit 3b8f934

Browse files
author
lixiang.2533
committed
add new solution
Change-Id: If6f7335b1074dfa253258862f63043df38d9a9b1
1 parent e1d94ba commit 3b8f934

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‎merge_md.md‎

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```
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import os
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from glob import glob
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dirs = os.listdir('/Users/lixiang.2533/Desktop/Blogs/Leetcode/')
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dir = '/Users/lixiang.2533/Desktop/Blogs/Leetcode'
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dirs = ['二分法']
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dirs = ['二分法','双指针','栈','排序','贪心']
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for a in dirs:
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md_list = glob(os.path.join(dir, a, '*.md'))
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md_list = sorted(md_list)

‎二分法总结.md‎

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### 0.二分法总结.md
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### 二分法总结
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- 找target
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- [ ] 69 x 的平方根:牛顿法
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- [ ] 74 搜索二维矩阵
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示例 1:
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输入:matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
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Tips
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依旧是二分法,强迫症执着的
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依旧是二分法,强迫症执着的使用左右闭区间,这个时候想要知道最后返回的是左还是右区间只要模拟一下推出的挺狂求可以
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### 287. 寻找重复数.md
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给定一个包含 n + 1 个整数的数组 nums ,其数字都在 [1, n] 范围内(包括 1 和 n),可知至少存在一个重复的整数。
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进阶:
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如何证明 nums 中至少存在一个重复的数字?
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```python
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class Solution:
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def search(self, nums: List[int], target: int) -> int:
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示例 1:
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输入:nums = [5,7,7,8,8,10], target = 8
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示例 1:
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输入:matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
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进阶:
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这是 搜索旋转排序数组 的延伸题目,本题中的 nums 可能包含重复元素。
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1. 在33题的基础上,因为数值可能存在重复,所以当左边界或者右边界和mid相同的时候都向内收缩一步再进行判断
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‎值得二刷.md‎

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‎动态规划/0.动态规划总结.md‎

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- [ ] 63 不同路径 II : 同上,碰到障碍保持0
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- [ ] 64 最小路径和:$ dp[i][j] = min(dp[i-1][j],dp[i][j-1])+ grid[i$][j]
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- [ ] 70 爬楼梯:$dp[n] = dp[n-1]+dp[n-2] ,ドル长度是n
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- [ ] 96 不同的二叉搜索树: $dp(n) += dp(j-1) * dp(n-j)$
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- [ ] 343 整数拆分: $dp[i] = max(dp[i-j]*j, (i-j)*j),ドル拆分or不拆分
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- [ ] 509 斐波那契数 : $dp[n] = dp[n-1]+dp[n-2] ,ドル 长度是n+1
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- [ ] 746 使用最小花费爬楼梯:$dp[i] = min(dp[i-2], dp[i-1]) + cost[i],ドル 最后返回$min(dp[-1],dp[-2])$

‎动态规划/343. 整数拆分.md‎

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dp[2] = 1
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for i in range(3,n+1):
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for j in range(1,i-1):
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dp[i] =max(dp[i], max(dp[i-j] * j, (i-j) * j))
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dp[i] =max(dp[i], dp[i-j] * j, (i-j) * j)
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print(dp)
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return dp[-1]
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```

‎树/96. 不同的二叉搜索树.md‎ renamed to ‎动态规划/96. 不同的二叉搜索树.md‎

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def numTrees(self, n: int) -> int:
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dp = [1] + [0] * n
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for i in range(1,n+1):
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for j in range(1,i+1):
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dp[i] += dp[j-1] * dp[i-j]
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for j in range(i):
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dp[i] += dp[j] * dp[i-j-1]
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return dp[-1]
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```
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