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"]
]

这个题目就是在每一次寻找完第一个回文串，然后在剩下的字符串中又是找到第一个回文串，一直这样到最后一个。

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

public class Solution131 {
    static LinkedList> ans = new LinkedList>();
    public List> partition(String s) {
        ans.clear();
        LinkedList list = new LinkedList();
        helper(s + "*", list, 0);
        return ans;
    }

    private void helper(String s, LinkedList list1, int n) {
        if (s.equals("*") || s.equals(" ") || s == null) {
            List list2 = new ArrayList();
            for (String string : list1) {
                list2.add(string);
            }
            ans.add(list2);
            return;
        }
        for (int i = 1; i < s.length(); i++) {
            String cur = s.substring(0, i);
            String res = s.substring(i, s.length());
            if (isPartitioning(cur)) {
                list1.add(cur);
                helper(res, list1, n + 1);
                list1.pollLast();
            }
        }
    }

    private boolean isPartitioning(String cur) {
        String cur1 = new StringBuffer(cur.trim()).reverse().toString();
        return cur.equals(cur1);
    }

    public static void main(String[] args) {
        new Solution131().partition("aab");
        System.out.println(ans);

    }

}

这个和分ip的是一样的