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

Author: kant-li

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