Leetcode 205. Isomorphic Strings

205. Isomorphic Strings

Total Accepted: 62899 Total Submissions: 206548 Difficulty: Easy

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.

思路：判断2个字符串是否同构。其实就是映射的问题，如果s和t的字符是一一映射（双射）的关系，就可以说s和t是同构的。

代码：

方法一：将s和t中的字符分别一一映射于整数集合，再对比映射以后的s和t是否相等。

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

方法二：
看是否能将s转换为t：

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        char m['z'+1]={0};
        bool b['z'+1]={false};
        int i;
        for(i=0;i<s.size();i++){
            if(!m[s[i]]&&!b[t[i]]){
                m[s[i]]=t[i];
                b[t[i]]=true;
            }
            if(m[s[i]]==t[i]){
                continue;
            }
            return false;
        }
        return true;
    }
};

方法三：看当前字符最近一次出现的位置是否相等。

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;
            //因为初始值为0，所以要i+1
            m1[s[i]] = i + 1; //m1记录s[i]在s中最近一次出现的位置
            m2[t[i]] = i + 1;
        }
        return true;
    }
};