Leetcode: Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.



Return all possible palindrome partitioning of s.



For example, given s = "aab",

Return



  [

    ["aa","b"],

    ["a","a","b"]

  ]

难度：98，与Word Break II问题一样，NP（枚举）与DP结合的问题。参考了网上的做法，这道题是求一个字符串中回文子串的切割，并且输出切割结果，其实是Word Break II和Longest Palindromic Substring结合，该做的我们都做过了。首先我们根据Longest Palindromic Substring中的方法建立一个字典，得到字符串中的任意子串是不是回文串的字典。接下来就跟Word Break II一样，根据字典的结果进行切割，然后按照循环处理递归子问题的方法，如果当前的子串满足回文条件，就递归处理字符串剩下的子串。如果到达终点就返回当前结果。算法的复杂度跟Word Break II一样，取决于结果的数量，最坏情况是指数量级的。

这里建立字典以方便查找String s中任意substring是不是回文，用到了DP，用法很精髓。

 1 public class Solution {

 2     public ArrayList<ArrayList<String>> partition(String s) {

 3         ArrayList<ArrayList<String>> partitions = new ArrayList<ArrayList<String>>();

 4         if (s == null || s.length() == 0) {

 5             return partitions;

 6         }

 7         boolean[][] dic = getdict(s);

 8         ArrayList<String> partition = new ArrayList<String>();

 9         helper(s, dic, 0, partition, partitions);

10         return partitions;

11     }

12     

13     public void helper(String s, boolean[][] dic, int starter, ArrayList<String> partition, ArrayList<ArrayList<String>> partitions) {

14         if (starter == s.length()) {

15             partitions.add(new ArrayList<String>(partition));

16             return;

17         }

18         for (int j=starter; j<s.length(); j++) {

19             if (dic[starter][j]) {

20                 partition.add(s.substring(starter, j+1));

21                 helper(s, dic, j+1, partition, partitions);

22                 partition.remove(partition.size() - 1);

23             }

24         }

25     }

26     

27     public boolean[][] getdict(String s) {

28         boolean[][] dic = new boolean[s.length()][s.length()];

29         for (int i=s.length()-1; i>=0; i--) {

30             for (int j=i; j<s.length(); j++) {

31                 if ((s.charAt(i) == s.charAt(j)) && ((j-i<2) || dic[i+1][j-1])) {

32                     dic[i][j] = true;

33                 }

34             }

35         }

36         return dic;

37     }

38 }