Leetcode--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：本題求全部解，通常使用深度優先+回溯的方法。對於求最優解或解的數量通常使用動態規劃的方法。

import java.util.*;
public class Solution {
 
        public ArrayList<ArrayList<String>> partition(String s) {
        if(s==null){
            return null;
        }
        ArrayList<ArrayList<String>> result=new ArrayList<ArrayList<String>>();
        ArrayList<String> cur=new ArrayList<String>();
        find(s,result,cur);
        return result;
    }

    public void find(String s,ArrayList<ArrayList<String>> result,ArrayList<String> cur){
        if(s.length()==0){
            //注意要new 一個ArrayList才能夠
            result.add(new ArrayList<>(cur));
            return;
        }
        for(int i=1;i<=s.length();i++){
            if(ispalindrome(s.substring(0,i))){
                cur.add(s.substring(0,i));
                //回溯家+遞歸
                find(s.substring(i,s.length()),result,cur);
                if(cur.size()!=0) {
                   cur.remove(cur.size() - 1);
                }
            }
        }
    }

    //頭尾字符依次比較，判斷是不是迴文字符串
    public boolean ispalindrome(String s){
        int i=0;
        int j=s.length()-1;
        while(i<=j){
            if(s.charAt(i)==s.charAt(j)){
                i++;
                j--;
                continue;
            }else{
                return false;
            }
        }
        return true;
    }
}