Leetcode: Triangle

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

clarification:  Here "adjacent" of a[i][j] means a[i+1][j] and a[i][j+1], if available(not out of bound), while a[i+1][j-1] is not "adjacent" element.
Note: it's not necessary to worry about the boundary of the index. The length of row i+1 is always bigger than that of row i by 1, so here j+1 is always valid.


This is a typical DP problem and can be solved with both top-down recursive method and bottom-up DP table.

My Solutions:

Top-down recursive solution:

    public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
        if(triangle.size()==0) return -1;
        int n = 0;
 
        return minimumTotal(triangle, 0, n);
    }
    public int minimumTotal(ArrayList<ArrayList<Integer>> triangle, int curLayer, int Index){
        
        int ret =0;
        int temp1=0, temp2=0;
        
        if(curLayer == triangle.size()) return 0;
        
        ArrayList<Integer> al = triangle.get(curLayer);
        
        
        temp1 = minimumTotal(triangle, curLayer+1, Index);
        
        temp2 = minimumTotal(triangle, curLayer+1, Index+1);
        
        ret = temp1>=temp2? al.get(Index)+temp2 : al.get(Index)+temp1;   
            
        return ret;
        
    }

This solution got TLE error in leetcode.

Bottom-up solution: (Note the O(n) extra space requirement, I used only an ArrayList<Integer> with the length equal to the last ArrayList from the input).漂亮！

public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
        if(triangle.size()==0) return -1;
        int n = triangle.size();

        
        ArrayList<Integer> ret = new ArrayList<Integer>();
        
        ArrayList<Integer> al = triangle.get(n-1);
        for(int j= 0; j< al.size(); j++){
            ret.add(al.get(j));
        }
        for(int i = n-2; i >= 0; i--){
            al = triangle.get(i);
            for(int j= 0; j< al.size(); j++){
                int temp = ret.get(j)<=ret.get(j+1)?ret.get(j) :ret.get(j+1);
                ret.set(j, al.get(j) + temp);
            }
            ret.remove(ret.size()-1);
        }
        
        return ret.get(0);    
        
    }

Pass the leetcode test!!