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;
}
};