Leetcode053--找到分支最小路径和

Posted 2022-04-16 劲火星空

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];