[LeetCode]Isomorphic Strings-优快云博客

本文链接：https://blog.youkuaiyun.com/CiaoLiang/article/details/45438345

本文探讨了如何通过映射关系判断两个字符串是否为同构词，详细解释了使用两个map进行一一对应匹配的方法，包括实例分析和代码实现。

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

Two strings 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.

For example,
Given "egg", "add", return true.

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:

You may assume both s and t have the same length.

[LeetCode Source]

分析：Isomorphic指的是同构词，即s和t中相同位置的词是一一对应的关系。比如“egg","add"。e对应a，g对应d。

注意这种对应关系是相互的。采用两个map就可以解决问题了。

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        map<char,char> contrast1;
        map<char,char> contrast2;
        for(int i=0; i<s.size(); ++i){
            if(contrast1.find(s[i])!=contrast1.end()){
                if(contrast1[s[i]]!=t[i]||contrast2[t[i]]!=s[i]){
                    return false;
                }
            }
            else {
                contrast1[s[i]] = t[i];
                if(contrast2.find(t[i])!=contrast2.end()){
                    if(contrast1[s[i]]!=t[i]||contrast2[t[i]]!=s[i]){
                    return false;
                 }
                }
                else contrast2[t[i]] = s[i];
            }
        }
        return true;
    }
};