LeetCode（205）Isomorphic Strings

题目

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.

分析

判断给定的两个字符串是否同构，字符串字面上很容易就能看得出来；

但是怎么用算法来判断呢？

仔细分析：
e <——> a
g <——> d
g <——> d
感觉像离散数学中的双向映射。

在数据结构中可以选择map，选择unordered_map来存储字母映射关系，它的搜索性能为常量。

AC代码

class Solution {public:    bool isIsomorphic(string s, string t) {        if (s.size() != t.size())            return false;        //求两个字符串的长度        int len = s.size();        //首先验证字符串s—>t的映射        unordered_map<char, char> um;        for (int i = 0; i < len; ++i)        {            auto pos = um.find(s[i]);            if (pos == um.end())                um.insert({ s[i], t[i] });            else{                if ((*pos).second != t[i])                    return false;            }//else        }//for        //再验证字符串t—>s的映射        um.clear();        for (int i = 0; i < len; ++i)        {            auto pos = um.find(t[i]);            if (pos == um.end())                um.insert({ t[i], s[i] });            else{                if ((*pos).second != s[i])                    return false;            }//else        }//for        return true;    }};

GitHub测试程序源码