267. 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.

For example:

Given s = "aabb", return ["abba", "baab"].

Given s = "abc", return [].

Hint:

1. If a palindromic permutation exists, we just need to generate the first half of the string.
2. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

链接： http://leetcode.com/problems/palindrome-permutation-ii/

题解：

求一个字符串能生成的all palindromic permutaitons。这个题目又写得很长。 总的来说我们要考虑以下一些点:

1. 给定s是否能生成Palindrome
2. 生成的Palindrome的奇偶性，是奇数的话中间的字符是哪一个
3. 利用permutation II的原理，计算左半部string
4. 根据Palindrome的奇偶性判断是否要加上中间的独立字符
5. 根据计算出的左半部string，计算右半部string，合并，加入到结果集中
6. 复杂度的计算(留给二刷了)

Time Complexity - O(2ⁿ)， Space Complexity - O(2ⁿ)

public class Solution {
    private boolean isOddPalindrome;
    private String singleChar;
    
    public List<String> generatePalindromes(String s) {
        List<String> res = new ArrayList<>();
        String evenPalindromeString = generateEvenPalindromeString(s);
        if(evenPalindromeString.length() == 0) {
            if(this.isOddPalindrome)
                res.add(singleChar);
            return res;
        }
        
        char[] arr = evenPalindromeString.toCharArray();
        Arrays.sort(arr);
        
        StringBuilder sb = new StringBuilder();
        boolean[] visited = new boolean[arr.length];
        generatePalindromes(res, sb, arr, visited);
        return res;
    }
    
    private void generatePalindromes(List<String> res, StringBuilder sb, char[] arr, boolean[] visited) {
        if(sb.length() == arr.length) {  // we only get permutation of left part of the result palindrome
            String s = sb.toString();
            res.add(this.isOddPalindrome ? (s + this.singleChar + reverse(s)) : (s + reverse(s)) );
            return;
        }
        
        for(int i = 0; i < arr.length; i++) {
            if(visited[i] || (i > 0 && arr[i] == arr[i - 1] && !visited[i - 1])) {      //skip duplicate
                continue;
            }
            if(!visited[i]) {
                visited[i] = true;
                sb.append(arr[i]);
                generatePalindromes(res, sb, arr, visited);
                sb.setLength(sb.length() - 1);
                visited[i] = false;
            }
        }
    }
    
    private String reverse(String s) {              //reverse string
        char[] arr = s.toCharArray();
        int lo = 0, hi = s.length() - 1;
        while(lo < hi) {
            char c = arr[lo];
            arr[lo++] = arr[hi];
            arr[hi--] = c;
        }
        return String.valueOf(arr);
    }
    
    private String generateEvenPalindromeString(String s) {         // get even chars of palindrome
        Set<Character> set = new HashSet<>();
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if(set.contains(c)) {
                set.remove(c);
                sb.append(c);               // append only even counted chars
            } else {
                set.add(c);
            }
        }
        
        if(set.size() <= 1) {
            if(set.size() == 1) {
                for(char chr : set) {
                    this.singleChar = String.valueOf(chr);
                }
                this.isOddPalindrome = true;
            }
            return sb.toString();
        } else {
            return "";
        }
    }
}

 

Reference:

https://leetcode.com/discuss/53626/ac-java-solution-with-explanation