feat: python solutions

* Fix solution on problem 88 with js, fixs azl397985856#62

* Fix solution on problem 88, fixs azl397985856#62

* Add python3 solution for problem 31

* Add python3 solution for problem 33

* Add python3 solution for problem 39

* Add python3 solution for problem 40

* Add language support description

problems/31.next-permutation.md

@@ -56,6 +56,8 @@ Here are some examples. Inputs are in the left-hand column and its corresponding
 - 找到从右边起`第一个大于nums[i]的`，并将其和nums[i]进行交换
 ## 代码
 
+* 语言支持: Javascript，Python3
+
 ```js
 /*
  * @lc app=leetcode id=31 lang=javascript

problems/33.search-in-rotated-sorted-array.md

@@ -62,7 +62,7 @@ Output: -1
 - 找出有序区间，然后根据target是否在有序区间舍弃一半元素
 ## 代码
 
-* 语言支持: Javascript
+* 语言支持: Javascript，Python3
 
 ```js
 /*
@@ -115,6 +115,38 @@ var search = function(nums, target) {
  return -1;
 };
 ```
+Python3 Code:
+```python
+class Solution:
+ def search(self, nums: List[int], target: int) -> int:
+ """用二分法，先判断左右两边哪一边是有序的，再判断是否在有序的列表之内"""
+ if len(nums) <= 0:
+ return -1
+
+ left = 0
+ right = len(nums) - 1
+ while left < right:
+ mid = (right - left) // 2 + left
+ if nums[mid] == target:
+ return mid
+
+ # 如果中间的值大于最左边的值，说明左边有序
+ if nums[mid] > nums[left]:
+ if nums[left] <= target <= nums[mid]:
+ right = mid
+ else:
+ # 这里 +1，因为上面是 <= 符号
+ left = mid + 1
+ # 否则右边有序
+ else:
+ # 注意：这里必须是 mid+1，因为根据我们的比较方式，mid属于左边的序列
+ if nums[mid+1] <= target <= nums[right]:
+ left = mid + 1
+ else:
+ right = mid
+
+ return left if nums[left] == target else -1
+```
 
 ## 扩展
 

problems/39.combination-sum.md

@@ -54,6 +54,8 @@ A solution set is:
 
 ## 代码
 
+* 语言支持: Javascript，Python3
+
 ```js
 /*
  * @lc app=leetcode id=39 lang=javascript
@@ -126,6 +128,45 @@ var combinationSum = function(candidates, target) {
  return list;
 };
 ```
+Python3 Code:
+```python
+class Solution:
+ def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+ """
+ 回溯法，层层递减，得到符合条件的路径就加入结果集中，超出则剪枝；
+ 主要是要注意一些细节，避免重复等；
+ """
+ size = len(candidates)
+ if size <= 0:
+ return []
+
+ # 先排序，便于后面剪枝
+ candidates.sort()
+
+ path = []
+ res = []
+ self._find_path(target, path, res, candidates, 0, size)
+
+ return res
+
+ def _find_path(self, target, path, res, candidates, begin, size):
+ """沿着路径往下走"""
+ if target == 0:
+ res.append(path.copy())
+ else:
+ for i in range(begin, size):
+ left_num = target - candidates[i]
+ # 如果剩余值为负数，说明超过了，剪枝
+ if left_num < 0:
+ break
+ # 否则把当前值加入路径
+ path.append(candidates[i])
+ # 为避免重复解，我们把比当前值小的参数也从下一次寻找中剔除，
+ # 因为根据他们得出的解一定在之前就找到过了
+ self._find_path(left_num, path, res, candidates, i, size)
+ # 记得把当前值移出路径，才能进入下一个值的路径
+ path.pop()
+```
 
 ## 相关题目
 

problems/40.combination-sum-ii.md

@@ -55,6 +55,8 @@ A solution set is:
 
 ## 代码
 
+* 语言支持: Javascript，Python3
+
 ```js
 /*
  * @lc app=leetcode id=40 lang=javascript
@@ -130,6 +132,44 @@ var combinationSum2 = function(candidates, target) {
  return list;
 };
 ```
+Python3 Code:
+```python
+class Solution:
+ def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
+ """
+ 与39题的区别是不能重用元素，而元素可能有重复；
+ 不能重用好解决，回溯的index往下一个就行；
+ 元素可能有重复，就让结果的去重麻烦一些；
+ """
+ size = len(candidates)
+ if size == 0:
+ return []
+
+ # 还是先排序，主要是方便去重
+ candidates.sort()
+
+ path = []
+ res = []
+ self._find_path(candidates, path, res, target, 0, size)
+
+ return res
+
+ def _find_path(self, candidates, path, res, target, begin, size):
+ if target == 0:
+ res.append(path.copy())
+ else:
+ for i in range(begin, size):
+ left_num = target - candidates[i]
+ if left_num < 0:
+ break
+ # 如果存在重复的元素，前一个元素已经遍历了后一个元素与之后元素组合的所有可能
+ if i > begin and candidates[i] == candidates[i-1]:
+ continue
+ path.append(candidates[i])
+ # 开始的 index 往后移了一格
+ self._find_path(candidates, path, res, left_num, i+1, size)
+ path.pop()
+```
 
 ## 相关题目