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feat: 增加Python Code

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`@@ -43,6 +43,8 @@ Output:`
`43`	`43`
`44`	`44`	`## 代码`
`45`	`45`
	`46`	`+* 语言支持: Javascript,Python3`
	`47`	`+`
`46`	`48`	```js
`47`	`49`	`/*`
`48`	`50`	`* @lc app=leetcode id=46 lang=javascript`
`@@ -94,6 +96,30 @@ var permute = function(nums) {`
`94`	`96`	`return list`
`95`	`97`	`};`
`96`	`98`	```
	`99`	`+Python3 Code:`
	`100`	+```Python
	`101`	`+class Solution:`
	`102`	`+ def permute(self, nums: List[int]) -> List[List[int]]:`
	`103`	`+ """itertools库内置了这个函数"""`
	`104`	`+ return itertools.permutations(nums)`
	`105`	`+`
	`106`	`+ def permute2(self, nums: List[int]) -> List[List[int]]:`
	`107`	`+ """自己写回溯法"""`
	`108`	`+ res = []`
	`109`	`+ def _backtrace(nums, pre_list):`
	`110`	`+ if len(nums) <= 0:`
	`111`	`+ res.append(pre_list)`
	`112`	`+ else:`
	`113`	`+ for i in nums:`
	`114`	`+ # 注意copy一份新的调用,否则无法正常循环`
	`115`	`+ p_list = pre_list.copy()`
	`116`	`+ p_list.append(i)`
	`117`	`+ left_nums = nums.copy()`
	`118`	`+ left_nums.remove(i)`
	`119`	`+ _backtrace(left_nums, p_list)`
	`120`	`+ _backtrace(nums, [])`
	`121`	`+ return res`
	`122`	+```
`97`	`123`
`98`	`124`	`## 相关题目`
`99`	`125`

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`@@ -40,6 +40,8 @@ Output:`
`40`	`40`
`41`	`41`	`## 代码`
`42`	`42`
	`43`	`+* 语言支持: Javascript,Python3`
	`44`	`+`
`43`	`45`	```js
`44`	`46`	`/*`
`45`	`47`	`* @lc app=leetcode id=47 lang=javascript`
`@@ -96,6 +98,34 @@ var permuteUnique = function(nums) {`
`96`	`98`	`return list;`
`97`	`99`	`};`
`98`	`100`	```
	`101`	`+Python3 code:`
	`102`	+```Python
	`103`	`+class Solution:`
	`104`	`+ def permuteUnique(self, nums: List[int]) -> List[List[int]]:`
	`105`	`+ """与46题一样,当然也可以直接调用itertools的函数,然后去重"""`
	`106`	`+ return list(set(itertools.permutations(nums)))`
	`107`	`+`
	`108`	`+ def permuteUnique(self, nums: List[int]) -> List[List[int]]:`
	`109`	`+ """自己写回溯法,与46题相比,需要去重"""`
	`110`	`+ # 排序是为了去重`
	`111`	`+ nums.sort()`
	`112`	`+ res = []`
	`113`	`+ def _backtrace(nums, pre_list):`
	`114`	`+ if len(nums) <= 0:`
	`115`	`+ res.append(pre_list)`
	`116`	`+ else:`
	`117`	`+ for i in range(len(nums)):`
	`118`	`+ # 如果是同样的数字,则之前一定已经生成了对应可能`
	`119`	`+ if i > 0 and nums[i] == nums[i-1]:`
	`120`	`+ continue`
	`121`	`+ p_list = pre_list.copy()`
	`122`	`+ p_list.append(nums[i])`
	`123`	`+ left_nums = nums.copy()`
	`124`	`+ left_nums.pop(i)`
	`125`	`+ _backtrace(left_nums, p_list)`
	`126`	`+ _backtrace(nums, [])`
	`127`	`+ return res`
	`128`	+```
`99`	`129`
`100`	`130`	`## 相关题目`
`101`	`131`

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`@@ -88,6 +88,8 @@ var rotate = function(matrix) {`
`88`	`88`
`89`	`89`	`## 代码`
`90`	`90`
	`91`	`+* 语言支持: Javascript,Python3`
	`92`	`+`
`91`	`93`	```js
`92`	`94`	`/*`
`93`	`95`	`* @lc app=leetcode id=48 lang=javascript`
`@@ -121,3 +123,26 @@ var rotate = function(matrix) {`
`121`	`123`	`}`
`122`	`124`	`};`
`123`	`125`	```
	`126`	`+Python3 Code:`
	`127`	+```Python
	`128`	`+class Solution:`
	`129`	`+ def rotate(self, matrix: List[List[int]]) -> None:`
	`130`	`+ """`
	`131`	`+ Do not return anything, modify matrix in-place instead.`
	`132`	`+ 先做矩阵转置(即沿着对角线翻转),然后每个列表翻转;`
	`133`	`+ """`
	`134`	`+ n = len(matrix)`
	`135`	`+ for i in range(n):`
	`136`	`+ for j in range(i, n):`
	`137`	`+ matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]`
	`138`	`+ for m in matrix:`
	`139`	`+ m.reverse()`
	`140`	`+`
	`141`	`+ def rotate2(self, matrix: List[List[int]]) -> None:`
	`142`	`+ """`
	`143`	`+ Do not return anything, modify matrix in-place instead.`
	`144`	`+ 通过内置函数zip,可以简单实现矩阵转置,下面的代码等于先整体翻转,后转置;`
	`145`	`+ 不过这种写法的空间复杂度其实是O(n);`
	`146`	`+ """`
	`147`	`+ matrix[:] = map(list, zip(*matrix[::-1]))`
	`148`	+```

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`@@ -68,6 +68,8 @@ var groupAnagrams = function(strs) {`
`68`	`68`
`69`	`69`	`## 代码`
`70`	`70`
	`71`	`+* 语言支持: Javascript,Python3`
	`72`	`+`
`71`	`73`	```js
`72`	`74`	`/*`
`73`	`75`	`* @lc app=leetcode id=49 lang=javascript`
`@@ -100,3 +102,21 @@ var groupAnagrams = function(strs) {`
`100`	`102`	`return Object.values(hashTable);`
`101`	`103`	`};`
`102`	`104`	```
	`105`	`+Python3 Code:`
	`106`	+```Python
	`107`	`+class Solution:`
	`108`	`+ def groupAnagrams(self, strs: List[str]) -> List[List[str]]:`
	`109`	`+ """`
	`110`	`+ 思路同上,在Python中,这里涉及到3个知识点:`
	`111`	`+ 1. 使用内置的 defaultdict 字典设置默认值;`
	`112`	`+ 2. 内置的 ord 函数,计算ASCII值(等于chr)或Unicode值(等于unichr);`
	`113`	`+ 3. 列表不可哈希,不能作为字典的键,因此这里转为元组;`
	`114`	`+ """`
	`115`	`+ str_dict = collections.defaultdict(list)`
	`116`	`+ for s in strs:`
	`117`	`+ s_key = [0] * 26`
	`118`	`+ for c in s:`
	`119`	`+ s_key[ord(c)-ord('a')] += 1`
	`120`	`+ str_dict[tuple(s_key)].append(s)`
	`121`	`+ return str_dict.values()`
	`122`	+```

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