[LeetCode112]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.

Analysis:

DP problem,
Transaction function: a[n][i] = a[n][i]+min(a[n-1][i], a[n-1][i-1]).
Note that in this problem, "adjacent" of a[i][j] means a[i-1][j] and a[i-1][j-1], if available(not out of bound), while a[i-1][j+1] is not "adjacent" element.

If we do this from up to down, it is complicated. While from down to up, we could use only one array to scan every row to get result

Java

public int minimumTotal(List<List<Integer>> triangle) {
        int n = triangle.size();
        if(n==0) return 0;
        int m = triangle.get(n-1).size();
        int [] result = new int[m];
        for(int i=n-1;i>=0;i--){
        	for(int j=0;j<triangle.get(i).size();j++){
        		if(i==n-1){
        			result[j] = triangle.get(i).get(j);
        			continue;
        		}
        		result[j] = Math.min(result[j], result[j+1])+triangle.get(i).get(j);
        	}
        }
        return result[0];
    }

c++

int minimumTotal(vector<vector<int> > &triangle) {
    int row = triangle.size();
    if(row == 0) return 0;
    vector<int> minV(triangle[row-1].size());
    for(int i = row-1; i>=0;i--){
        int col = triangle[i].size();
        for(int j=0; j<col;j++){
            if(i==row-1){
                minV[j] = triangle[i][j];
                continue;
            }
            minV[j] = min(minV[j],minV[j+1])+triangle[i][j];
        }
    }
    return minV[0];
    }