LeetCode #389. Find the Difference

题目

389. Find the Difference

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解题方法

先统计两个字符串中每个字母的出现次数，记为dic_s和dic_t，先遍历dic_s，找一个在dic_t中没出现的字母，或者在dic_t中出现了但是频数和dic_s中不一样的字母，找到了就break不做了，要是没找到就再去dic_t中找和dic_s中不一样的字母，最后返回。
时间复杂度：O(n)
空间复杂度：O(n)

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代码

class Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        dic_s = collections.Counter(s)
        dic_t = collections.Counter(t)
        
        rat = None
        for key in dic_s.keys():
            if key not in dic_t or dic_t[key] != dic_s[key]:
                rat = key
                break
        
        if not rat:
            for key in dic_t.keys():
                if key not in dic_s or dic_s[key] != dic_t[key]:
                    rat = key
                    break
        return rat