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242. 有效的字母异位词

English Version

题目描述

给定两个字符串 s 和 t ，编写一个函数来判断 t 是否是 s 的字母异位词。

注意：若 s 和 t 中每个字符出现的次数都相同，则称 s 和 t 互为字母异位词。

 

示例 1:

输入: s = "anagram", t = "nagaram"
输出: true

示例 2:

输入: s = "rat", t = "car"
输出: false

 

提示:

• 1 <= s.length, t.length <= 5 * 104
• s 和 t 仅包含小写字母

 

进阶: 如果输入字符串包含 unicode 字符怎么办？你能否调整你的解法来应对这种情况？

解法

数组或哈希表累加 s 中每个字符出现的次数，再减去 t 中对应的每个字符出现的次数。遍历结束后，若字符中出现次数不为 0 的情况，返回 false，否则返回 true。

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;
    }
}

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);
};

TypeScript

function isAnagram(s: string, t: string): boolean {
    const n = s.length;
    const m = t.length;
    return n === m && [...s].sort().join('') === [...t].sort().join('');
}

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

Rust

impl Solution {
    pub fn is_anagram(s: String, t: String) -> bool {
        let n = s.len();
        let m = t.len();
        if n != m {
            return false;
        }
        let mut s = s.chars().collect::<Vec<char>>();
        let mut t = t.chars().collect::<Vec<char>>();
        s.sort();
        t.sort();
        for i in 0..n {
            if s[i] != t[i] {
                return false;
            }
        }
        true
    }
}

impl Solution {
    pub fn is_anagram(s: String, t: String) -> bool {
        let n = s.len();
        let m = t.len();
        if n != m {
            return false;
        }
        let (s, t) = (s.as_bytes(), t.as_bytes());
        let mut count = [0; 26];
        for i in 0..n {
            count[(s[i] - b'a') as usize] += 1;
            count[(t[i] - b'a') as usize] -= 1;
        }
        count.iter().all(|&c| c == 0)
    }
}

C

int cmp(const void *a, const void *b) {
    return *(char *) a - *(char *) b;
}

bool isAnagram(char *s, char *t) {
    int n = strlen(s);
    int m = strlen(t);
    if (n != m) {
        return 0;
    }
    qsort(s, n, sizeof(char), cmp);
    qsort(t, n, sizeof(char), cmp);
    return !strcmp(s, t);
}

bool isAnagram(char *s, char *t) {
    int n = strlen(s);
    int m = strlen(t);
    if (n != m) {
        return 0;
    }
    int count[26] = {0};
    for (int i = 0; i < n; i++) {
        count[s[i] - 'a']++;
        count[t[i] - 'a']--;
    }
    for (int i = 0; i < 26; i++) {
        if (count[i]) {
            return 0;
        }
    }
    return 1;
}

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