LeetCode 131 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"]
  ]

思路：1.推断字符串的字串S.subString(i,j) [i<=j]是否为为回文子串,用boolean型的二维数组isPalindrome来存储该结果。在这个地方用了点小技巧，isPalindrome[i]j]会依赖于isPalindrome[i+1]j-1]  [i+2<=j].

          2.如果在求String s的全部回文子串，我们已经知道了s.substring(0,0),s.substring(0,1),s.substring(0,2),s.substring(0,3),....,s.substring(0,s.length()-2)的全部回文字串，那我们仅仅须要遍历isPalindrome[i]js.lenth()-1]是不是true｛即从s.substring(i)是不是回文字符串｝若为true，我们取出s.substring(i-1)的全部回文字串，并在每一种可能性末尾加入s.substring(i)就可以，代码例如以下

public class Solution {

	public boolean[][] isPalindrome(String s) {
		boolean[][] isPalindrome = new boolean[s.length()][s.length()];
		for (int i = 0; i < s.length(); i++)
			isPalindrome[i][i] = true;

		for (int i = 0; i < s.length() - 1; i++)
			isPalindrome[i][i + 1] = (s.charAt(i) == s.charAt(i + 1));

		for (int length = 2; length < s.length(); length++) {
			for (int start = 0; start + length < s.length(); start++) {
				isPalindrome[start][start + length] = isPalindrome[start + 1][start
						+ length - 1]
						&& s.charAt(start) == s.charAt(start + length);
			}
		}
		return isPalindrome;
	}

	public List<List<String>> partition(String s) {
		boolean[][] isPalindrome= isPalindrome(s);
		HashMap<Integer,List<List<String>>> hm=new HashMap<Integer,List<List<String>>>();
		for(int i=0;i<s.length();i++){
			List<List<String>> ls=new ArrayList<List<String>>();
			if(isPalindrome[0][i]){
				ArrayList<String> temp=new ArrayList<String>();
				temp.add(s.substring(0, i+1));
				ls.add(temp);
			}
			
			for(int j=1;j<=i;j++){
				if(isPalindrome[j][i]){
					List<List<String>> l=hm.get(j-1);
					List<List<String>> al=new ArrayList<List<String>>();
					for(List<String> temp:l){
						ArrayList<String> clone=new ArrayList<String>(temp);
						clone.add(s.substring(j, i+1));
						al.add(clone);
					}
					ls.addAll(al);
				}	
			}
			hm.put(i,ls);
		}
		return hm.get(s.length()-1);
	}
}