[LeetCode] 1400. Construct K Palindrome Strings

Posted 2021-06-12 CNoodle

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.

Example 1:

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"

Example 2:

Input: s = "leetcode", k = 3
Output: false
Explanation: It is impossible to construct 3 palindromes using all the characters of s.

Example 3:

Input: s = "true", k = 4
Output: true
Explanation: The only possible solution is to put each character in a separate string.

Example 4:

Input: s = "yzyzyzyzyzyzyzy", k = 2
Output: true
Explanation: Simply you can put all z\'s in one string and all y\'s in the other string. Both strings will be palindrome.

Example 5:

Input: s = "cr", k = 7
Output: false
Explanation: We don\'t have enough characters in s to construct 7 palindromes.

Constraints:

• 1 <= s.length <= 10^5
• All characters in s are lower-case English letters.
• 1 <= k <= 10^5

构造 K 个回文字符串。

给你一个字符串 s 和一个整数 k 。请你用 s 字符串中 所有字符 构造 k 个非空 回文串 。

如果你可以用 s 中所有字符构造 k 个回文字符串，那么请你返回 True ，否则返回 False 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/construct-k-palindrome-strings
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是贪心。题意是试图把 input 字符串 s 分割，请你判断是否有可能把 input 字符串分割成 k 份，同时每份子串都是一个回文串。这里贪心的思想是，如果我可以把 s 分成 k 个回文串的话，那么出现次数为奇数的字母个数一定不能超过 k 个，否则就不行。出现次数为 3 次或者 5 次的字母的确可以自成一个回文，但是他不能再和其他出现次数为奇数的字母合并成另一个回文了，所以出现次数为奇数的字母有多少，就会产生多少个子串。如果这个数字大于 k，就不能满足题意。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public boolean canConstruct(String s, int k) {
 3         int odd = 0;
 4         int len = s.length();
 5         int[] count = new int[26];
 6         for (int i = 0; i < len; i++) {
 7             count[s.charAt(i) - \'a\']++;
 8         }
 9         for (int j = 0; j < count.length; j++) {
10             if (count[j] % 2 != 0) {
11                 odd++;
12             }
13         }
14         return k <= len && odd <= k;
15     }
16 }

 

LeetCode 题目总结