LeetCode 267. 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.