leetcode 205. Isomorphic Strings 同构字符串判断 + HashMap

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;    }};