LeetCode 242. Valid Anagram

原题链接在这里：https://leetcode.com/problems/valid-anagram/

题目：

Given two strings s and t, write a function to determine if t is an anagram of s.

For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.

Note:
You may assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?

题解：

维护一个map, s对应char位++, t对应char位--. 若是map有非零位便return false.

Time Complexity: O(n), n 为string长度.

Space: O(1). 用了一个array当map.

AC Java：

Method 1:

public class Solution {
    public boolean isAnagram(String s, String t) {
        if(s == null && t == null){
            return true;
        }
        if(s == null || t == null){
            return false;
        }
        if(s.length() != t.length()){
            return false;
        }
        
        int [] map = new int[256];
        for(int i = 0; i<s.length(); i++){
            map[s.charAt(i)]++;
        }
        for(int i = 0; i<t.length(); i++){
            map[t.charAt(i)]--;
        }
        for(int i = 0; i<256; i++){
            if(map[i] != 0){
                return false;
            }
        }
        return true;
    }
}

这道题可以直接sort 两个string 然后比较，相同就返回true.

不过要注意for Strings, equals() method works as str1.equals(str2). for Arrays, equals() method works as Arrays.equals(arr1, arr2).

Time Compleixty: O(nlogn). 有sort在内.

Space: O(n), string换成了char array.

Method 2:

public class Solution {
    public boolean isAnagram(String s, String t) {
         //Method 2
        char[] temp1 = s.toCharArray();
        char[] temp2 = t.toCharArray();
        Arrays.sort(temp1);
        Arrays.sort(temp2);
        
        return Arrays.equals(temp1,temp2);
    }
}

跟上Find All Anagrams in a String, Group Anagrams.