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 = '\0';
         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 != '\0') {
                 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;
     }
 }