Leetcode刷题记录——面试题 01.02. 判定是否互为字符重排

我们先用一个字典储存s1中的全部字符和出现次数
再遍历s2，从字典中的对应位置减

class Solution:
    def CheckPermutation(self, s1: str, s2: str) -> bool:
        s1dict = {}
        for letter in s1:
            if letter in s1dict:
                s1dict[letter] += 1
            else:
                s1dict[letter] = 1
        for letter in s2:
            if letter not in s1dict or s1dict[letter] == 0:
                return False
            else:
                s1dict[letter] -= 1
        
        return True