leetcode- Isomorphic Strings
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.
方法二图例:
直接看代码分析:
package leetcode;
import java.util.*;
public class IsomorphicStrings {
// 题目:All occurrences of a character....
// 框架相同,即只要在某位置上出现相同字符,那么另一个单词相应位置也一定出现相同字符;
// 像 palepr、 titlte这种情况,就返回false,可用相同字母替换的;
public static boolean isIsomorphic(String sString, String tString) {
/*
* 方法一:直接用map if(s==null && t==null) return true;
* if(s.length()!=t.length()) return false; //用键值对保存两个字符串的相应字符!good idea
* HashMap<Character,Character> map = new HashMap<>(); int i=0;
* while(i<s.length()){ //前面要分析,如果键值相同要怎么办?那么如果key相同,而value不同,就返回false
* if(map.containsKey(s.charAt(i))){ char tmp = map.get(s.charAt(i));
* if(tmp!=t.charAt(i)) return false; }else
* if(map.containsValue(t.charAt(i))) return false; else{
* //只有确保key不相同的时候才可以往里放键值对,所以放值一定是最后的步骤(当键不相同的时候才可以放)
* map.put(s.charAt(i), t.charAt(i)); } i++; System.out.println(map); }
*
* return true;
*/
/*
* 方法二:这里是用两个数组去模拟HashMap
* 新建两个s,t数组,其下标为自己当前的元素,但是存的值是对方的相应位置上的元素,建立起了映射关系
*
*/
char[] s = sString.toCharArray();
char[] t = tString.toCharArray();
int length = s.length;
if (length != t.length)
return false;
char[] sm = new char[256];
char[] tm = new char[256];
for (int i = 0; i < length; i++) {
char sc = s[i];
char tc = t[i];
if (sm[sc] == 0 && tm[tc] == 0) {
sm[sc] = tc; // 关键idea!
tm[tc] = sc;
} else {
if (sm[sc] != tc || tm[tc] != sc) {
return false;
}
}
}
return true;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.print(isIsomorphic("papepr", "titlte"));
}
}