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面试题 01.02. 判定是否互为字符重排

English Version

题目描述

给定两个字符串 s1 和 s2，请编写一个程序，确定其中一个字符串的字符重新排列后，能否变成另一个字符串。

示例 1：

输入: s1 = "abc", s2 = "bca"
输出: true

示例 2：

输入: s1 = "abc", s2 = "bad"
输出: false

说明：

• 0 <= len(s1) <= 100
• 0 <= len(s2) <= 100

解法

方法一：

使用哈希表作为字符计数器，O(n) 时间内解决。

方法二：

按照字符编码重新排序字符串，再检查两者一致性。

Python3

class Solution:
    def CheckPermutation(self, s1: str, s2: str) -> bool:
        n1, n2 = len(s1), len(s2)
        if n1 != n2:
            return False
        counter = Counter()
        for i in range(n1):
            counter[s1[i]] += 1
            counter[s2[i]] -= 1
        return all(v == 0 for v in counter.values())

Java

class Solution {
    public boolean CheckPermutation(String s1, String s2) {
        int n1 = s1.length();
        int n2 = s2.length();
        if (n1 != n2) {
            return false;
        }
        int[] counter = new int[128];
        for (int i = 0; i < n1; ++i) {
            ++counter[s1.charAt(i)];
            --counter[s2.charAt(i)];
        }
        for (int v : counter) {
            if (v != 0) {
                return false;
            }
        }
        return true;
    }
}

JavaScript

var CheckPermutation = function (s1, s2) {
    let n1 = s1.length,
        n2 = s2.length;
    if (n1 != n2) return false;
    let counter = {};
    for (let i = 0; i < n1; i++) {
        let cur1 = s1.charAt(i),
            cur2 = s2.charAt(i);
        counter[cur1] = (counter[cur1] || 0) + 1;
        counter[cur2] = (counter[cur2] || 0) - 1;
    }
    return Object.values(counter).every(v => v == 0);
};

Go

func CheckPermutation(s1 string, s2 string) bool {
	freq := make(map[rune]int)
	for _, r := range s1 {
		freq[r]++
	}
	for _, r := range s2 {
		if freq[r] == 0 {
			return false
		}
		freq[r]--
	}
	for _, v := range freq {
		if v != 0 {
			return false
		}
	}
	return true
}

C++

class Solution {
public:
    bool CheckPermutation(string s1, string s2) {
        int n1 = s1.size();
        int n2 = s2.size();
        if (n1 != n2) return 0;
        vector<int> counter(128);
        for (int i = 0; i < n1; ++i)
        {
            ++counter[s1[i]];
            --counter[s2[i]];
        }
        for (int v : counter)
            if (v) return 0;
        return 1;
    }
};

TypeScript

function CheckPermutation(s1: string, s2: string): boolean {
    const n = s1.length;
    const m = s2.length;
    if (n !== m) {
        return false;
    }
    const map = new Map<string, number>();
    for (let i = 0; i < n; i++) {
        map.set(s1[i], (map.get(s1[i]) || 0) + 1);
        map.set(s2[i], (map.get(s2[i]) || 0) - 1);
    }
    for (const v of map.values()) {
        if (v !== 0) {
            return false;
        }
    }
    return true;
}

function CheckPermutation(s1: string, s2: string): boolean {
    if (s1.length !== s2.length) {
        return false;
    }
    return (
        s1
            .split('')
            .sort((a, b) => a.charCodeAt(0) - b.charCodeAt(0))
            .join('') ===
        s2
            .split('')
            .sort((a, b) => a.charCodeAt(0) - b.charCodeAt(0))
            .join('')
    );
}

Rust

use std::collections::HashMap;

impl Solution {
    pub fn check_permutation(s1: String, s2: String) -> bool {
        let n = s1.len();
        let m = s2.len();
        if n != m {
            return false;
        }
        let s1: Vec<char> = s1.chars().collect();
        let s2: Vec<char> = s2.chars().collect();
        let mut map = HashMap::new();
        for i in 0..n {
            map.insert(s1[i], map.get(&s1[i]).unwrap_or(&0) + 1);
            map.insert(s2[i], map.get(&s2[i]).unwrap_or(&0) - 1);
        }
        map.values().all(|i| *i == 0)
    }
}

impl Solution {
    pub fn check_permutation(s1: String, s2: String) -> bool {
        let mut v1: Vec<char> = s1.chars().collect();
        let mut v2: Vec<char> = s2.chars().collect();
        v1.sort();
        v2.sort();
        v1 == v2
    }
}

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