LeetCode: palindrome-partitioning

题目描述

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"]
  ]

import java.util.*;
public class Partition {
	public static void main(String[] args) {
		
	}
	public ArrayList> PartionMain(String s){
		
		ArrayList> re=new ArrayList>();
		ArrayList list = new ArrayList();  
        
        if (s == null || s.length() == 0)  return re;
        
        calResult(re,list,s); 
		return re;
	}
	
	 /** 
     * 回溯 
     * @param arr : 最终要的结果集 ArrayList> 
     * @param list : 当前已经加入的集合 ArrayList 
     * @param s : 当前要处理的字符串 
     */  
	private void  calResult(ArrayList> arr,ArrayList list,String s) {
		 //当处理到传入的字符串长度等于0,则这个集合list满足条件，加入到结果中  
		 if (s.length() == 0)  
	            arr.add(new ArrayList(list));  
        int len = s.length();  
        //递归调用  
        //字符串由前往后，先判断str.substring(0, i)是否是回文字符串  
        //如果是的话，继续调用函数calResult，把str.substring(i)字符串剪切传入做处理  
        for (int i=1; i<=len; ++i){  
            String subStr = s.substring(0, i);  
            if (isPalindrome(subStr)){  
                list.add(subStr);  
                String restSubStr = s.substring(i);  
                calResult(arr,list,restSubStr);  
                list.remove(list.size()-1); //每次将刚才加进来的移出去
            }  
        }  
	}
	/** 
     * 判断一个字符串是否是回文字符串 
     */  
    private boolean isPalindrome(String str){  
          
        int i = 0;  
        int j = str.length() - 1;  
        while (i < j){  
            if (str.charAt(i) != str.charAt(j)){  
                return false;  
            }  
            i++;  
            j--;  
        }  
        return true;  
    }  
	
}