【Leetcode】【240402】205. Isomorphic Strings(map的使用方法摘要）

新年度开始了，开始真剑努力毕业+就职。
定个小目标：每日一题
多谢鸡桑提醒国际服早上刷新，让我在半夜打卡成功，很好很贴心
算了以后还是放github吧，两边搬怪麻烦的

Description

Given two strings s and t, determine if they are isomorphic.

Two strings s and t are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.

Example 1

Input: s = "egg", t = "add"
Output: true

Example 2

Input: s = "foo", t = "bar"
Output: false

Example 3

Input: s = "paper", t = "title"
Output: true

Constraints

1 <= s.length <= 5 * $10^4$
t.length == s.length
s and t consist of any valid ascii character.

Solution

对于很久没写代码且只会暴力的我来说也是很简单的一道题。
思路大概是：首先将s和t的字符对应，发现同一个字符的第二种对应返回false，无事终了再将t和s重复上述步骤。无事终了返回true。
需要注意的一点是：判断是否出现过该字符的对应时，使用了map.count()
错误解法：判断是否为空（由于不知道map初始化无法用’ ‘判断
直接来看代码段：

class Solution {
public:
    bool isIsomorphic(string s, string t) 
    {
        map<char,char>m;
        map<char,char>n;
        for(int i=0;i<s.length();++i)
        {
            if(m.count(s[i]))
            {
                if(m[s[i]]!=t[i]) return false;
            }
            else m[s[i]]=t[i];
        }
        for(int i=0;i<t.length();++i)
        {
            if(n.count(t[i]))
            {
                if(n[t[i]]!=s[i]) return false;
            }
            else n[t[i]]=s[i];
        }
        return true;
    }
};

附录

map的使用方法摘录