README_EN.md

242. Valid Anagram

中文文档

Description

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

 

Example 1:

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

Example 2:

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

 

Constraints:

• 1 <= s.length, t.length <= 5 * 104
• s and t consist of lowercase English letters.

 

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

Solutions

Python3

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        chars = [0] * 26
        for i in range(len(s)):
            chars[ord(s[i]) - ord('a')] += 1
            chars[ord(t[i]) - ord('a')] -= 1
        return all(c == 0 for c in chars)

Java

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        int[] chars = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++chars[s.charAt(i) - 'a'];
            --chars[t.charAt(i) - 'a'];
        }
        for (int c : chars) {
            if (c != 0) {
                return false;
            }
        }
        return true;
    }
}

TypeScript

function isAnagram(s: string, t: string): boolean {
    if (s.length != t.length) return false;
    let record = new Array(26).fill(0);
    let base = 'a'.charCodeAt(0);
    for (let i = 0; i < s.length; ++i) {
        ++record[s.charCodeAt(i) - base];
        --record[t.charCodeAt(i) - base];
    }
    return record.every(v => v == 0);
};

C++

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.size() != t.size())
            return false;
        vector<int> chars(26, 0);
        for (int i = 0, n = s.size(); i < n; ++i)
        {
            ++chars[s[i] - 'a'];
            --chars[t[i] - 'a'];
        }
        for (int c : chars)
        {
            if (c != 0)
                return false;
        }
        return true;
    }
};

Go

func isAnagram(s string, t string) bool {
	if len(s) != len(t) {
		return false
	}
	var chars [26]int
	for i := 0; i < len(s); i++ {
		chars[s[i]-'a']++
		chars[t[i]-'a']--
	}
	for _, c := range chars {
		if c != 0 {
			return false
		}
	}
	return true
}

JavaScript

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isAnagram = function(s, t) {
    if (s.length != t.length) return false;
    let record = new Array(26).fill(0);
    let base = 'a'.charCodeAt(0);
    for (let i = 0; i < s.length; ++i) {
        ++record[s.charCodeAt(i) - base];
        --record[t.charCodeAt(i) - base];
    }
    return record.every(v => v == 0);
};

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