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add some k-sum code
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‎Array/K-SUM题解.md

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### 题目详解
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#### 1.[1-Two Sum](https://leetcode.com/problems/two-sum/)[easy]
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#K-SUM
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#### 2.[167-Two Sum II](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/)[easy]
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##1.Two Sum
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- 题目:1 Two sum 【easy】
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#### 3.[170.Two Sum III - Data structure design](https://leetcode.com/problems/two-sum-iii-data-structure-design/)[easy]
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- 解法:hash-map
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- code
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#### 4.[653. Two Sum IV - Input is a BST](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/)[easy]
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Python
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```python
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#hash-map
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class Solution(object):
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def twoSum(self, nums, target):
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"""
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:type nums: List[int]
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:type target: int
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:rtype: List[int]
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"""
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if len(nums) <= 1:
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return False
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keys = {}
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for i, v in enumerate(nums):
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if target-v in keys:
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return [keys[target-v],i]
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else:
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keys[v] = i
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return []
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```
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## 2.Two Sum II - Input array is sorted
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- 题目:167 Two Sum II - Input array is sorted 【easy】
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- 解法:
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- hash-map:排序了而已,和第一题很像啊,就是把下标改一下
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- 双指针:与二分法结合
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- 二分查找:
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- code:
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Java
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```java
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/*hash-map*/
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class Solution {
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public int[] twoSum(int[] numbers, int target) {
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int[] result = new int[2];
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HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
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for (int i = 0;i<numbers.length;i++){
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if (map.containsKey(target-numbers[i])){
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result[0]=map.get(target-numbers[i]);
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result[1]= i+1;
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}
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map.put(numbers[i],i+1);
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}
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return result;
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}
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}
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```
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Python
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```python
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# 双指针法
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def twoSum1(self, numbers, target):
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l, r = 0, len(numbers)-1
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while l < r:
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s = numbers[l] + numbers[r]
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if s == target:
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return [l+1, r+1]
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elif s < target:
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l += 1
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else:
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r -= 1
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# dictionary
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def twoSum2(self, numbers, target):
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dic = {}
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for i, num in enumerate(numbers):
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if target-num in dic:
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return [dic[target-num]+1, i+1]
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dic[num] = i
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# binary search
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def twoSum(self, numbers, target):
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for i in xrange(len(numbers)):
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l, r = i+1, len(numbers)-1
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tmp = target - numbers[i]
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while l <= r:
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mid = l + (r-l)//2
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if numbers[mid] == tmp:
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return [i+1, mid+1]
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elif numbers[mid] < tmp:
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l = mid+1
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else:
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r = mid-1
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```
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‎Array/array.md

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## 数组
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数组
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### 什么是数组
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【一维数组】
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#### 1.K-Sum
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​ 这类题目通常会给定一个数组和一个值,让求出这个数组中两个/三个/K个值的和等于这个给定的值target。leetcode第一题就是two-sum,对于这类题目,首先看题目要求的时间复杂度和空间复杂度是什么,其次看有没有限制条件,如要求不能有重复的子数组或者要求按照升序/降序排列等。解法如下:
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- 暴力解法:最常见,但是通常会超时,只能作为备选,
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- hash-map:建立一个hash-map循环遍历一次即可
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- two-pointers:定位两个指针根绝和的大小来移动另外一个。这里设定的指针个数根据题目中K的个数来定。3Sum中可以设定3个指针,固定两个,移动另一个
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> 1.暴力解法:最常见,但是通常会超时,只能作为备选,
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>
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> 2.hash-map:建立一个hash-map循环遍历一次即可
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>
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> 3.two-pointers:定位两个指针根绝和的大小来移动另外一个。这里设定的指针个数根据题目中K的个数来定。3Sum中可以设定3个指针,固定两个,移动另一个。
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- 例题:167 Two Sum II - Input array is sorted 【easy】
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- 题解:
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- test case:
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- 解题思路:
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- code:
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Java
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```java
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/*hash-map*/
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class Solution {
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public int[] twoSum(int[] numbers, int target) {
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int[] result = new int[2];
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HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
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for (int i = 0;i<numbers.length;i++){
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if (map.containsKey(target-numbers[i])){
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result[0]=map.get(target-numbers[i]);
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result[1]= i+1;
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}
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map.put(numbers[i],i+1);
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}
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return result;
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}
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}
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```
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Python
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```python
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# 双指针法
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def twoSum1(self, numbers, target):
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l, r = 0, len(numbers)-1
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while l < r:
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s = numbers[l] + numbers[r]
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if s == target:
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return [l+1, r+1]
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elif s < target:
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l += 1
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else:
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r -= 1
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# dictionary
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def twoSum2(self, numbers, target):
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dic = {}
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for i, num in enumerate(numbers):
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if target-num in dic:
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return [dic[target-num]+1, i+1]
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dic[num] = i
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# binary search
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def twoSum(self, numbers, target):
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for i in xrange(len(numbers)):
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l, r = i+1, len(numbers)-1
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tmp = target - numbers[i]
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while l <= r:
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mid = l + (r-l)//2
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if numbers[mid] == tmp:
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return [i+1, mid+1]
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elif numbers[mid] < tmp:
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l = mid+1
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else:
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r = mid-1
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```
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#### 2.区间问题
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>解释:满足子数组和=7的最小长度数组是[4,3],所以output=2
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- 解题思路:求的数字要大于等于这个数字target,譬如这个testcase中,[2,3,1,2,4,3],从前往后相加,前四项相加的和为8.已经大于7了,但是我们的target是7,后面的几项也都是正整数,继续往后走,肯定也会大于target7,所以这个时候我们把left指针往右移动一位,那么就是相当于成为了一个滑动窗口(sliding window),这种方法在String类型的题目出现的更多。
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​ 在本题中,刚好最后两位4,3相加等于target7,所以滑动窗口滑到最后一位的时候满足题解,但是有个问题,如果在滑动的过程中,中间出现某个数字大于target或者等于target
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![image](https://raw.githubusercontent.com/huxiaoman7/leetcodebook/master/Array/pic/subarray.png)
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- code:
@@ -138,9 +211,62 @@ def minSubArrayLen(self, s, nums):
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#### Rotate题型
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- 例题:48. Rotate Image 【medium】
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- 题解:
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- test case:
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> Given input matrix =
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> [
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> [1,2,3],
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> [4,5,6],
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> [7,8,9]
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> ],
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>
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> rotate the input matrix in-place such that it becomes:
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> [
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> [7,4,1],
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> [8,5,2],
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> [9,6,3]
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> ]
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- 解题方法:
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- 1.转置后再翻转
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- 2.旋转四个角
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- code:
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Python
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```python
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# 旋转后再翻转
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class solution:
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def rotate(self,matrix):
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"""
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:type matrix: List[List[int]]
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:rtype: void Do not return anything, modify matrix in-place instead.
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"""
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n = len(matrix[0])
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# 转置
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for i in range(n):
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for j in range(i,n):
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matrix[j][i],matrix[i][j] = matrix[i][j],matrix[j][i]
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for i in range(n):
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matrix[i].reverse()
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# 旋转四个角
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```
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- 复杂度分析:
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- 转置后翻转:时间$O(N^2)$ ,空间$O(1)$.
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-
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