【字符串】B032 构造 K 个回文字符串（利用性质）

一、题目描述

Given a string s and an integer k. You should construct k non-empty palindrome strings using all the characters in s.

Return True if you can use all the characters in s to construct k palindrome strings or False otherwise.

Input: s = "annabelle", k = 2
Output: true
Explanation: You can construct two palindromes using all characters in s.
Some possible constructions "anna" + "elble", "anbna" + "elle", "anellena" + "b"

方法一：考察回文串的特性

• 对于单个字符串来说：当出现次数为奇数的字符 >=2 个，就构不成回文串了，即 odd 需要满足 odd <= 1。
• 所以当要构成 k 个回文串时，奇数字符必须 odd <= k 个，即至少不大于 k 个。
  • 需满足：k <= s.length();

class Solution {
    public boolean canConstruct(String s, int k) {
        if (s.length() < k)
            return false;
        char[] cs = s.toCharArray();
        int odd = 0, n = s.length(), m[] = new int[26];
        for (char c : cs)
            m[c-'a']++;
        for (int i : m) if ((i & 1) == 1) {
            odd++;
        }
        return odd <= k;
    }
}

复杂度分析

• 时间复杂度：$O(N)$，
• 空间复杂度：$O(1)$，