LeetCode Palindrome Permutation II

原题链接在这里：https://leetcode.com/problems/palindrome-permutation-ii/

题目：

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

Example 1:

Input: "aabb"
Output: ["abba", "baab"]

Example 2:

Input: "abc"
Output: []

题解：

先判断是否palindrome, 若不是返回空的res.

若是，先判断是否有一个奇数位，若有，改char放中间. 然后两边分别加.

Note: check to find single char after updating the map.

Don't forget to minus its count by 1 after appending it to item.

Time Complexity: O(2^n). n = s.length().

Space: O(n). stack space.

AC Java:

 class Solution {
     public List<String> generatePalindromes(String s) {
         List<String> res = new ArrayList<>();
         if(s == null || s.length() == 0){
             return res;
         }

         int [] count = new int[256];
         int n = s.length();
         int po = -1;
         for(int i = 0; i<n; i++){
             count[s.charAt(i)]++;
         }

         int oneCount = 0;
         for(int i = 0; i<256; i++){
             oneCount += count[i]%2;

             if(count[i]%2 == 1){
                 po = i;
             }
         }

         if(oneCount > 1){
             return res;
         }

         String init = "";
         if(po != -1){
             init += (char)po;
             count[po]--;
         }

         dfs(count, init, n, res);
         return res;
     }

     private void dfs(int [] count, String cur, int n, List<String> res){
         if(cur.length() == n){
             res.add(cur);
             return;
         }

         for(int i = 0; i<count.length; i++){
             if(count[i] > 0){
                 count[i] -= 2;
                 dfs(count, (char)i+cur+(char)i, n, res);
                 count[i] += 2;
             }
         }
     }
 }

与Palindrome Permutation相似.