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
sare 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 }