题解 | #字符串的排列#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param str string字符串 
# @return string字符串一维数组
#
class Solution:
    def Permutation(self , str: str) -> List[str]:
        # write code here
        ans = []
        path = [''] * len(str)
        onPath = [False] * len(str)
        str = list(str)
        str.sort()
        def dfs(i: int) -> None:
            if i == len(str):
                ans.append(path.copy())
                return
            for j in range(0, len(str)):
                if onPath[j] or (j > 0 and str[j] == str[j - 1] and not onPath[j - 1]):
                    continue
                path.append(str[j])
                onPath[j] = True
                dfs(i + 1)
                path.pop()
                onPath[j] = False
        dfs(0)
        return [''.join(item) for item in ans]