**Palindrome Partitioning

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题目：

https://leetcode.com/problems/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"]  ]

分析：

http://blog.csdn.net/linhuanmars/article/details/22777711

这里getDcit方法与上一篇中的dp数组的求解方法完全一样啊！！！

public class Solution {public List<List<String>> partition(String s) {    List<List<String>> res = new ArrayList<List<String>>();    if(s==null || s.length()==0)        return res;    helper(s, getDict(s),0,new ArrayList<String>(), res);    return res;}private void helper(String s, boolean[][] dict, int start, ArrayList<String> item, List<List<String>> res){    if(start==s.length())    {        res.add(new ArrayList<String>(item));        return;    }    for(int i=start;i<s.length();i++)    {        if(dict[start][i])        {            item.add(s.substring(start,i+1));            helper(s,dict,i+1,item,res);            item.remove(item.size()-1);        }    }}private boolean[][] getDict(String s){    boolean[][] dict = new boolean[s.length()][s.length()];    for(int i=s.length()-1;i>=0;i--)    {        for(int j=i;j<s.length();j++)        {            if(s.charAt(i)==s.charAt(j) && ((j-i<2)||dict[i+1][j-1]))            {                dict[i][j] = true;            }        }    }    return dict;}}

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