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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 }
Leetcode: Palindrome Partitioning
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原文地址：http://www.cnblogs.com/EdwardLiu/p/3980588.html
