[LeetCode] 242. Valid Anagram

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

Example 1:

Input: s = "anagram", t = "nagaram"
Output: true

Example 2:

Input: s = "rat", t = "car"
Output: 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?

有效的字母异位词。题意是给两个string，请判断他们是否互为字母异位词。

这题用counting sort做，会用到hashmap。思路是建立一个hashmap，先扫描s，记录s里面每个字母出现的次数；然后遍历t的时候，减去t里面每个字符出现的次数，如果这个过程中发现t中有任何字符在hashmap里不存在，就return false。若没有，则再次遍历hashmap，如果有任何一个key的value不是0，也return false。

时间O(n)

空间O(n)

Java实现

class Solution {
    public boolean isAnagram(String s, String t) {
        // corner case
        if (s.length() != t.length()) {
            return false;
        }

        // normal case
        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++;
            count[t.charAt(i) - 'a']--;
        }
        for (int j = 0; j < count.length; j++) {
            if (count[j] != 0) {
                return false;
            }
        }
        return true;
    }
}

JavaScript实现

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isAnagram = function(s, t) {
    let map = {};
    for (let i = 0; i < s.length; i++) {
        if (!map[s[i]]) {
            map[s[i]] = 1;
        } else {
            map[s[i]]++;
        }
    }
    for (let i = 0; i < t.length; i++) {
        if (!map[t[i]] || map[t[i]] < 0) {
            return false;
        } else {
            map[t[i]]--;
        }
    }
    for (let key in map) {
        if (map[key] !== 0) {
            return false;
        }
    }
    return true;
};

LeetCode 题目总结