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.

这道题考察的是同构字符串的判断,使用HashMap统一编码即可。

代码如下:

import java.util.HashMap;
import java.util.Map;

/*
 * 我这里借助map来完成统一的映射,
 * 只需要遍历一次即可
 * */
public class Solution
{
    public boolean isIsomorphic(String s, String t)
    {
        if(s==null || t==null)
            return false;
        if(s.length()!=t.length())
            return false;

        Map<Character, Integer> map1=new HashMap<Character, Integer>();
        Map<Character, Integer> map2=new HashMap<Character, Integer>();
        int counta=0,countb=0;
        StringBuilder a=new StringBuilder();
        StringBuilder b=new StringBuilder();

        for(int i=0;i<s.length();i++)
        {
            if(map1.containsKey(s.charAt(i)))
                a.append(map1.get(s.charAt(i)));
            else
            {
                map1.put(s.charAt(i), counta);
                a.append(counta);
                counta++;
            }

            if(map2.containsKey(t.charAt(i)))
                b.append(map2.get(t.charAt(i)));
            else
            {
                map2.put(t.charAt(i), countb);
                b.append(countb);
                countb++;
            }

            //而这需要同步,所以这里加了一个小小的判断
            if(counta!=countb)
                return false;
        }
        if(a.toString().equals(b.toString()))
            return true;
        else
            return false;
    }
}

下面是C++的做法,就是做统一编码,做一次遍历即可解决问题

代码如下:

#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;


class Solution 
{
public:
    bool isIsomorphic(string s, string t) 
    {
        if (s.length() != t.length())
            return false;
        map<char,int> mp1, mp2;
        int key1 = 0, key2 = 0;
        string res1 = "", res2 = "";
        for (int i = 0; i < s.length(); i++)
        {
            if (mp1.find(s[i]) != mp1.end() )
                res1 += to_string(mp1[s[i]]);
            else
            {
                mp1[s[i]] = key1;
                res1 += to_string(key1++);
            }

            if (mp2.find(t[i]) != mp2.end())
                res2 += to_string(mp2[t[i]]);
            else
            {
                mp2[t[i]] = key2;
                res2 += to_string(key2++);
            }
            if (key1 != key2)
                return false;
        }
        return res1 == res2;
    }
};