README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0200-0299/0266.Palindrome%20Permutation/README_EN.md	Bit Manipulation Hash Table String

266. Palindrome Permutation 🔒

中文文档

Description

Given a string s, return true if a permutation of the string could form a palindrome and false otherwise.

 

Example 1:

Input: s = "code"
Output: false

Example 2:

Input: s = "aab"
Output: true

Example 3:

Input: s = "carerac"
Output: true

 

Constraints:

• 1 <= s.length <= 5000
• s consists of only lowercase English letters.

Solutions

Solution 1

Python3

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

Java

class Solution {
    public boolean canPermutePalindrome(String s) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            ++cnt[c - 'a'];
        }
        int odd = 0;
        for (int x : cnt) {
            odd += x & 1;
        }
        return odd < 2;
    }
}

C++

class Solution {
public:
    bool canPermutePalindrome(string s) {
        vector<int> cnt(26);
        for (char& c : s) {
            ++cnt[c - 'a'];
        }
        int odd = 0;
        for (int x : cnt) {
            odd += x & 1;
        }
        return odd < 2;
    }
};

Go

func canPermutePalindrome(s string) bool {
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	odd := 0
	for _, x := range cnt {
		odd += x & 1
	}
	return odd < 2
}

TypeScript

function canPermutePalindrome(s: string): boolean {
    const cnt: number[] = new Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0) - 97];
    }
    return cnt.filter(c => c % 2 === 1).length < 2;
}

JavaScript

/**
 * @param {string} s
 * @return {boolean}
 */
var canPermutePalindrome = function (s) {
    const cnt = new Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt() - 'a'.charCodeAt()];
    }
    return cnt.filter(c => c % 2 === 1).length < 2;
};