Palindrome Permutation II 解答

Question

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.

Answer

这道题思考起来其实不难，就是步骤比较多，要小心。

1. 判断输入是否能有permutation构成palindrome => HashMap或HashSet，这里因为判断完后还有下一步，所以采用Map

2. 根据Map获得first half of the string

3. Backtracking生成first half的permutation

4. 根据生成的permutation，补齐last half

另外解答中处理反转String的方法也值得学习：先利用原String构造一个StringBuffer，然后从后往前遍历原String，将char一一加到StringBuffer中。

public class Solution {
    public List<String> generatePalindromes(String s) {
        List<String> result = new ArrayList<>();
        if (s == null || s.length() == 0) {
            return result;
        }
        int len = s.length();
        Map<Character, Integer> map = new HashMap<>();
        if (!isPalindromePermutation(map, s)) {
            return result;
        }
        char mid = '';
        StringBuilder sb = new StringBuilder();
        for (char cur : map.keySet()) {
            int num = map.get(cur);
            while (num > 1) {
                sb.append(cur);
                num -= 2;
            }
            if (num == 1) {
                mid = cur;
            }
        }
        Set<String> prefix = new HashSet<String>();
        boolean[] visited = new boolean[sb.length()];
        dfs(sb, visited, prefix, "");
        for (String left : prefix) {
            StringBuffer tmp = new StringBuffer(left);
            if (mid != '') {
                tmp.append(mid);
            }
             
            for (int i = left.length() - 1; i >= 0; i--) {
                tmp.append(left.charAt(i));
            }
            result.add(tmp.toString());
        }
        return result;
    }
    
    private void dfs(StringBuilder sb, boolean[] visited, Set<String> result, String prev) {
        if (prev.length() == sb.length()) {
            result.add(prev);
            return;
        }
        int len = sb.length();
        int prevIndex = -1;
        for (int i = 0; i < len; i++) {
            if (visited[i]) {
                continue;
            }
            if (prevIndex != -1 && sb.charAt(i) == sb.charAt(prevIndex)) {
                continue;
            }
            visited[i] = true;
            dfs(sb, visited, result, prev + sb.charAt(i));
            visited[i] = false;
            prevIndex = i;
        }
    }
    
    private boolean isPalindromePermutation(Map<Character, Integer> map, String s) {
        int tolerance = 0;
        int len = s.length();
        for (int i = 0; i < len; i++) {
            char cur = s.charAt(i);
            if (!map.containsKey(cur)) {
                map.put(cur, 1);
            } else {
                map.put(cur, map.get(cur) + 1);
            }
        }
        for (char cur : map.keySet()) {
            int num = map.get(cur);
            if (num % 2 == 1) {
                tolerance++;
            }
        }
        return tolerance < 2;
    }
}