[Leetcode] Isomorphic Strings

描述

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.

分析1

判断两个字符串是否是同构字符串，即判断两个字符串的字符是否是一一对应的关系。
因此，我们需要用两个哈希表，将每个字符串的字符与其位置信息对应起来，并通过位置信息将两个哈希表联系起来。

对于两个字符串的同一位置的两个字符：
1. 如果都没出现在哈希表中：分别添加进哈希表，将位置作为哈希表的值；
2. 只有一个出现在哈希表中：必然不同构；
3. 都出现在哈希表中：比较两个哈希表中它们的值是否相同。

代码1

class Solution {public:    bool isIsomorphic(string s, string t) {        unordered_map<char,int> m1,m2;        for (int i = 0; i < s.size(); i++) {            bool in_m1 = m1.find(s[i]) != m1.end();            bool in_m2 = m2.find(t[i]) != m2.end();            if ((in_m1 && !in_m2) || (!in_m1 && in_m2)) return false;            if (in_m1 && in_m2 && m1[s[i]] != m2[t[i]]) return false;            if (!in_m1 && !in_m2) {m1[s[i]] = i + 1; m2[t[i]] = i + 1;}        }        return true;    }};

分析2

当然也可以用数组简化代码。

代码2

class Solution {public:    bool isIsomorphic(string s, string t) {        int m1[256] = {0}, m2[256] = {0}, n = s.size();        for (int i = 0; i < n; i++){            if (m1[s[i]] != m2[t[i]]) return false;            m1[s[i]] = i + 1; m2[t[i]] = i + 1;        }        return true;    }};