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
        path = []

        n = len(str)

        L2 = list(str)
        ans = []
        L4 = []
        visited = [False for i in range(n)]
	#要记得排序。排序了才可以用到那个剪短枝条的公式
        L2.sort()

        def MyPermutation(L2, L, ans):
            if len(L) == n:
                ans.append(path.copy())
                return

            else:
                for i in range(len(L2)):
                    if visited[i]:
                        continue
                    if i>0 and L2[i-1]==L2[i] and visited[i-1]==False:
                        continue

                    path.append(L2[i])
                    visited[i] = True
                    MyPermutation(L2, L, ans)
                    path.pop()
                    visited[i] = False

        MyPermutation(L2, path, ans)
        for x in ans:
            L4.append("".join(x))
        return L4