1. 题目描述
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.
给定两个字符串,判断两个字符串是不是同构的。
2. 解题思路
这个题目和 [290] Word Pattern很相似,只不过290题将一个字符串看做一个模式,判断同构性,而本题是讲一个字符看做一个模式判断同构性。本次的解法和290题略有不同,但是基本思路都是一样的。需要两个数据结构,一个数据结构去保存从s到t中的字符的映射关系,另一个数据结构去保存是否这个已经被映射过,如s为aab,而t为ccc,他们的模式是不同的,因为a已经映射到c,所以b不能再被映射到c。
3. Code
public class Solution {
public boolean isIsomorphic(String s, String t) {
// character是否只是ascii字符?
int [] maps = new int[256]; // s中的值映射到t中的哪一个
boolean [] isMapped = new boolean[256]; // 是否已经被map了
for(int i = 0; i < s.length(); ++i)
{
int svalue = (int)s.charAt(i);
int mapPos = maps[svalue]; // mapp的字符位置
// 如果不是初始值0,这个判断舍去了ascii码的第一个null,实测不影响
if(mapPos != 0)
{
if(mapPos != (int)t.charAt(i)) return false;
}else{
int tvalue = (int)t.charAt(i);
if(isMapped[tvalue]) return false;
maps[svalue] = tvalue;
isMapped[tvalue] = true;
}
}
return true;
}
}