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Commit 870b0ad

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kant-liazl397985856
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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
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‎problems/31.next-permutation.md‎

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

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

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@@ -62,7 +62,7 @@ Output: -1
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- 找出有序区间,然后根据target是否在有序区间舍弃一半元素
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## 代码
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* 语言支持: Javascript
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* 语言支持: Javascript,Python3
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```js
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/*
@@ -115,6 +115,38 @@ var search = function(nums, target) {
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return -1;
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};
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```
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Python3 Code:
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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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"""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"""
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if len(nums) <= 0:
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return -1
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left = 0
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right = len(nums) - 1
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while left < right:
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mid = (right - left) // 2 + left
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if nums[mid] == target:
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return mid
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# 如果中间的值大于最左边的值,说明左边有序
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if nums[mid] > nums[left]:
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if nums[left] <= target <= nums[mid]:
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right = mid
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else:
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# 这里 +1,因为上面是 <= 符号
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left = mid + 1
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# 否则右边有序
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else:
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# 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
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if nums[mid+1] <= target <= nums[right]:
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left = mid + 1
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else:
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right = mid
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return left if nums[left] == target else -1
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```
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## 扩展
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‎problems/39.combination-sum.md‎

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## 代码
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* 语言支持: Javascript,Python3
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```js
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/*
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* @lc app=leetcode id=39 lang=javascript
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return list;
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};
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```
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Python3 Code:
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```python
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class Solution:
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def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
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"""
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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;
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主要是要注意一些细节,避免重复等;
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"""
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size = len(candidates)
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if size <= 0:
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return []
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# 先排序,便于后面剪枝
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candidates.sort()
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path = []
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res = []
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self._find_path(target, path, res, candidates, 0, size)
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return res
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def _find_path(self, target, path, res, candidates, begin, size):
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"""沿着路径往下走"""
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if target == 0:
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res.append(path.copy())
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else:
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for i in range(begin, size):
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left_num = target - candidates[i]
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# 如果剩余值为负数,说明超过了,剪枝
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if left_num < 0:
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break
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# 否则把当前值加入路径
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path.append(candidates[i])
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# 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,
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# 因为根据他们得出的解一定在之前就找到过了
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self._find_path(left_num, path, res, candidates, i, size)
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# 记得把当前值移出路径,才能进入下一个值的路径
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path.pop()
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```
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## 相关题目
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‎problems/40.combination-sum-ii.md‎

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## 代码
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* 语言支持: Javascript,Python3
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```js
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/*
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* @lc app=leetcode id=40 lang=javascript
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return list;
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};
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```
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Python3 Code:
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```python
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class Solution:
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def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
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"""
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与39题的区别是不能重用元素,而元素可能有重复;
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不能重用好解决,回溯的index往下一个就行;
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元素可能有重复,就让结果的去重麻烦一些;
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"""
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size = len(candidates)
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if size == 0:
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return []
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# 还是先排序,主要是方便去重
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candidates.sort()
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path = []
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res = []
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self._find_path(candidates, path, res, target, 0, size)
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return res
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def _find_path(self, candidates, path, res, target, begin, size):
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if target == 0:
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res.append(path.copy())
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else:
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for i in range(begin, size):
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left_num = target - candidates[i]
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if left_num < 0:
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break
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# 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
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if i > begin and candidates[i] == candidates[i-1]:
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continue
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path.append(candidates[i])
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# 开始的 index 往后移了一格
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self._find_path(candidates, path, res, left_num, i+1, size)
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path.pop()
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```
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## 相关题目
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