Leetcode053--找到分支最小路径和

一、原题

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 is11(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.

二、中文

求最小值，从上到下的分支可以任意选择

三、举例

比如上面的分支的最小值就是2 + 3 + 5 + 1 = 11

四、思路

还是动态规划的思想，只不过这次是从下往上开始的，使用的是二维数组

五、程序

import java.util.*;public class Solution {    public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {        //也是通过动态规划的方式        int row = triangle.size();        int col = triangle.get(row-1).size();                int d[][] = new int[row][col];        //先将最后一行填满        for(int i = 0; i < col; i++){            d[row-1][i] = triangle.get(row-1).get(i);        }                //从下往上选取最小的依次填充        for(int i = row-2; i >= 0; i--){            //获得一行            ArrayList<Integer> l = triangle.get(i);            for(int j = 0;j<l.size();j++){                d[i][j] = l.get(j) + Math.min(d[i+1][j], d[i+1][j+1]);            }        }        return d[0][0];    }}