[LeetCode]--131. Palindrome Partitioning(backTracking && DFS && DP)

1. Problem

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"]]

2. Solution

2.1. BackTracking DFS Thought

利用backTracking()的方法, 不断进行回溯处理。并且进行DFS处理

public class Solution {    public List<List<String>> partition(String s) {       List<List<String>> res = new ArrayList<List<String>>();       List<String> list = new ArrayList<String>();       dfs(s,0,list,res);       return res;    }    public void dfs(String s, int pos, List<String> list, List<List<String>> res){        if(pos==s.length()) res.add(new ArrayList<String>(list));        else{            for(int i=pos;i<s.length();i++){                if(isPal(s,pos,i)){                    list.add(s.substring(pos,i+1));                    dfs(s,i+1,list,res);                    list.remove(list.size()-1);                }            }        }    }    public boolean isPal(String s, int low, int high){        while(low<high) if(s.charAt(low++)!=s.charAt(high--)) return false;        return true;    }}

2.2. DP Solution

public class Solution {    public static List<List<String>> partition(String s) {        int len = s.length();        List<List<String>>[] result = new List[len + 1];        result[0] = new ArrayList<List<String>>();        result[0].add(new ArrayList<String>());        boolean[][] pair = new boolean[len][len];        for (int i = 0; i < s.length(); i++) {            result[i + 1] = new ArrayList<List<String>>();            for (int left = 0; left <= i; left++) {                if (s.charAt(left) == s.charAt(i) && (i-left <= 1 || pair[left + 1][i - 1])) {                    pair[left][i] = true;                    String str = s.substring(left, i + 1);                    for (List<String> r : result[left]) {                        List<String> ri = new ArrayList<String>(r);                        ri.add(str);                        result[i + 1].add(ri);                    }                }            }        }        return result[len];    }}

Here the pair is to mark a range for the substring is a Pal. if pair[i][j] is true, that means sub string from i to j is pal.

The result[i], is to store from beginng until current index i (Non inclusive), all possible partitions. From the past result we can determine current result.

Reference

1. https://discuss.leetcode.com/topic/6186/java-backtracking-solution
2. https://discuss.leetcode.com/topic/2884/my-java-dp-only-solution-without-recursion-o-n-2/3
3. https://discuss.leetcode.com/category/139/palindrome-partitioning