| 标题: |
Palindrome Partitioning |
| 通过率： | 26.3% |
| 难度： | 中等 |
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"]
]
本题还是一个递归的过程，只是再递归的时候要判断是否为回文函数，然后就是字符串的截取，具体看代码
1 public class Solution { 2 public ArrayList<ArrayList<String>> partition(String s) { 3 ArrayList<ArrayList<String>> res=new ArrayList<ArrayList<String>>(); 4 ArrayList<String> tmp=new ArrayList<String>(); 5 getpa(res,tmp,s); 6 return res; 7 } 8 public void getpa(ArrayList<ArrayList<String>> res,ArrayList<String> tmp,String s){ 9 if(s.length()==0||s==null){ 10 res.add(new ArrayList<String>(tmp)); 11 } 12 else{ 13 for(int i=1;i<=s.length();i++){ 14 if(ispa(s.substring(0,i))){ 15 tmp.add(s.substring(0,i)); 16 getpa(res,tmp,s.substring(i)); 17 tmp.remove(tmp.size()-1); 18 } 19 } 20 } 21 } 22 public boolean ispa(String s){ 23 for(int i=0,j=s.length()-1;i<j;i++,j--){ 24 if(s.charAt(i)!=s.charAt(j)) 25 return false; 26 } 27 return true; 28 } 29 }
leetcode------Palindrome Partitioning
原文：http://www.cnblogs.com/pkuYang/p/4341941.html