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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

用一个哈希表作为字符计数器，O(n) 时间内解决。

Python3

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

Java

class Solution {
    public boolean CheckPermutation(String s1, String s2) {
        int n1 = s1.length(), n2 = s2.length();
        if (n1 != n2) {
            return false;
        }
        Map<Character, Integer> counter = new HashMap<>();
        for (int i = 0; i < n1; ++i) {
            char c1 = s1.charAt(i), c2 = s2.charAt(i);
            counter.put(c1, counter.getOrDefault(c1, 0) + 1);
            counter.put(c2, counter.getOrDefault(c2, 0) - 1);
        }
        for (int val : counter.values()) {
            if (val != 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
}

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