README_EN.md

01.04. Palindrome Permutation

中文文档

Description

Given a string, write a function to check if it is a permutation of a palin­ drome. A palindrome is a word or phrase that is the same forwards and backwards. A permutation is a rearrangement of letters. The palindrome does not need to be limited to just dictionary words.

 

Example1:

Input: "tactcoa"

Output: true（permutations: "tacocat"、"atcocta", etc.）

Solutions

Python3

class Solution:
    def canPermutePalindrome(self, s: str) -> bool:
        counter = Counter(s)
        return sum(1 for v in counter.values() if v % 2 == 1) <= 1

Java

class Solution {
    public boolean canPermutePalindrome(String s) {
        Map<Character, Integer> counter = new HashMap<>();
        for (char c : s.toCharArray()) {
            counter.put(c, counter.getOrDefault(c, 0) + 1);
        }
        int cnt = 0;
        for (int v : counter.values()) {
            cnt += v % 2;
        }
        return cnt < 2;
    }
}

Go

func canPermutePalindrome(s string) bool {
	m := make(map[rune]bool)
	count := 0
	for _, r := range s {
		if m[r] {
			m[r] = false
			count--
		} else {
			m[r] = true
			count++
		}
	}
	return count <= 1
}

C++

class Solution {
public:
    bool canPermutePalindrome(string s) {
        unordered_map<char, int> counter;
        for (char c : s) ++counter[c];
        int cnt = 0;
        for (auto& [k, v] : counter) cnt += v % 2;
        return cnt < 2;
    }
};

TypeScript

function canPermutePalindrome(s: string): boolean {
    const set = new Set<string>();
    for (const c of s) {
        if (set.has(c)) {
            set.delete(c);
        } else {
            set.add(c);
        }
    }
    return set.size <= 1;
}

Rust

use std::collections::HashSet;

impl Solution {
    pub fn can_permute_palindrome(s: String) -> bool {
        let mut set = HashSet::new();
        for c in s.chars() {
            if set.contains(&c) {
                set.remove(&c);
            } else {
                set.insert(c);
            }
        }
        set.len() <= 1
    }
}

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