有重復(fù)字符的全排列
整整30min 。糾結(jié)的點(diǎn)在于如何轉(zhuǎn)換為list，終止條件

# -*- coding:utf-8 -*-
class Solution:
    def Permutation(self, ss):
        s=list(ss)
        s.sort()
        tmp=self.permu(s)
        res=[]
        for c in tmp:
            res.append("".join(c)) #如何將結(jié)果轉(zhuǎn)換為字符串
        return res


    def permu(self,s):
        if len(s)==1:return [s]  #遞歸的終止條件出錯(cuò)
        res=[]
        for i in range(len(s)):
            if i==0 or s[i]!=s[i-1]:
                for p in self.permu(s[:i]+s[i+1:]):
                    res.append([s[i]]+p)
        return res

借鑒的是原來(lái)的有些奇葩的思路

# -*- coding:utf-8 -*-
class Solution:
    def Permutation(self, ss):
        if not ss: return []  # 初始條件直接輸入空字符串
        s=list(ss)
        s.sort()
        tmp=self.permu(s)
        res=[]
        for c in tmp:
            res.append("".join(c)) #如何將結(jié)果轉(zhuǎn)換為字符串
        return res


    def permu(self,s):
        # if len(s)==1:return [s]  #遞歸的終止條件出錯(cuò)
        res=[]
        for i in range(len(s)):
            if i==0 or s[i]!=s[i-1]:
                for p in self.permu(s[:i]+s[i+1:]):
                    res.append([s[i]]+p)
        return res or [[]] # 當(dāng)s為空時(shí)的巧妙的遞歸條件