Leetcode.205 Isomorphic Strings（Java）

Leetcode.205 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.

Example 1:

Input: s = "egg", t = "add"
Output: true

Example 2:

Input: s = "foo", t = "bar"
Output: false

Example 3:

Input: s = "paper", t = "title"
Output: true

Note:
You may assume both *s* and *t* have the same length.

Solution

双hashmap

两个都是字符串，作比较的是单个字符，如果未经处理直接存在一个表中可能会产生冲突

“ab”
"aa"

因此先考虑存入两个表（防止互相干扰）

class Solution {
    /*
    time:O(n)
    space:O(n)
    */
    public boolean isIsomorphic(String s, String t) {
        int[] sIndex = new int[256];
        int[] tIndex = new int[256];
        if(s.length() != t.length()){
            return false;
        }
        for(int i=1;i<s.length()+1;i++){
            //"aee" "aee"
            if(sIndex[s.charAt(i-1)] != tIndex[t.charAt(i-1)]){
                return false;
            }
            sIndex[s.charAt(i-1)] = i;
            tIndex[t.charAt(i-1)] = i;
        }
        return true;
        
    }
}

代码17,18行由于初始化是0，所以索引从1开始，防止0冲突

单hashmap

class Solution {
    /*
    time:O(n)
    space:O(n)
    */
    public boolean isIsomorphic(String s, String t) {
        int[] index = new int[512];
        if(s.length() != t.length()){
            return false;
        }
        for(int i=0;i<s.length();i++){
            //"aee" "aee"
            if(index[s.charAt(i)] != index[t.charAt(i)+256]){
                return false;
            }
            index[s.charAt(i)] = i+1;
            index[t.charAt(i)+256]=i+1;
        }
        return true;
        
    }
}