Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example 1:
[[1,3,1], [1,5,1], [4,2,1]]
Given the above grid map, return 7
. Because the path 1→3→1→1→1 minimizes the sum.
写这道题是因为这题的点赞率特别高,题目大意是给定一个数组,判断数组左上角到右下角的和最小是多少。
读完题就知道肯定是动态规划,而且比较简单,设定了一个大小和grid相同的数组dia,状态方程为dia[i][j]=grid[i][j]+Math.min(dia[i-1][j],dia[i][j-1])
因为只能向下或者向右移动,所以只需判断上面或左边最小值为多少,然后加一加即可。
下面是java代码。
class Solution {
public int minPathSum(int[][] grid) {
int m=grid.length;
int n=grid[0].length;
int [][]dia=new int[m][n];
dia[0][0]=grid[0][0];
for(int i=1;i<m;i++)
dia[i][0]=grid[i][0]+dia[i-1][0];
for(int i=1;i<n;i++)
dia[0][i]=grid[0][i]+dia[0][i-1];
//初始化完成,写状态方程
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
{
dia[i][j]=grid[i][j]+Math.min(dia[i-1][j],dia[i][j-1]);
}
return dia[m-1][n-1];
}
}
ps:一开始就给出grid长宽多好。