Leetcode Everyday: 266. Palindrome Permutation

原创于 2016-05-11 22:57:52 发布 · 353 阅读

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本文介绍了一种算法，用于判断给定的字符串是否可以通过重新排列形成回文串。核心思路是检查字符串中每个字符出现次数的奇偶性，确保至多只有一个字符的出现次数为奇数。

https://leetcode.com/problems/palindrome-permutation/

Given a string, determine if a permutation of the string could form a palindrome.

For example,

"code" -> False, "aab" -> True, "carerac" -> True.

public class Solution {
    public boolean canPermutePalindrome(String s) {
        char[] c = s.toCharArray();
        int[] i = new int[128];
        for(char cc: c){
            i[(int) cc]+=1;
        }
        boolean flag = true;
        for(int ii: i){
            if(ii%2!=0){
                if(flag) flag=false;
                else return false;
            }
        }
        
        return true;
        
    }
}

Only at most one character's occurrence can be odd. Suppose charset is ASCII