Palindrome Partitioning

https://leetcode.com/problems/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"]

  ]

解题思路：

这种题目一看就是DFS的题，中间用到一个 Valid Palindrome 的方法，判断一个string是不是palindrome。

对于每个位置start，取字符串的可能都是从start开始一直到结尾，所以要判断它是不是palindrome，是的话就加入current。对于下面的字符串继续DFS。到结尾了，回溯。

public class Solution {

    public List<List<String>> partition(String s) {

        List<List<String>> result = new LinkedList<List<String>>();

        if(s.length() == 0) {

            return result;

        }

        dfs(s, result, new LinkedList<String>(), 0);

        return result;

    }

    

    public void dfs(String s, List<List<String>> result, LinkedList<String> current, int start) {

        if(current.size() > 0 && !isPalindrome(current.getLast())) {

            return;

        }

        if(start == s.length()) {

            result.add(new LinkedList(current));

        }

        for(int i = 1; i < s.length() - start + 1; i++) {

            current.add(s.substring(start, start + i));

            dfs(s, result, current, start + i);

            current.remove(current.size() - 1);

        }

        

    }

    

    public boolean isPalindrome(String s) {

        if(s.length() == 0) {

            return true;

        }

        int start = 0, end = s.length() - 1;

        while(start < end) {

            if(s.charAt(start) != s.charAt(end)){

                return false;

            }

            start++;

            end--;

        }

        return true;

    }

}