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"]
  ]
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);
	}
}LeetCode 131 Palindrome Partitioning
原文地址:http://blog.csdn.net/mlweixiao/article/details/40507675