[LeetCode]64. Minimum Path Sum

Minimum Path Sum

问题描述

代码

代码分析

问题描述

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.

代码

class Solution {
public:
int a[500][500]={0};//到a[i][j]的最短距离
int minPathSum(vector<vector<int>>& grid) {
int w = grid.size();
int h = grid[0].size();
a[0][0] = grid[0][0];
for(int i = 0; i<w;++i)
for(int j = 0; j<h;++j)
if(!(i == 0 && j == 0))
if( i == 0)
a[i][j] = a[i][j-1] + grid[i][j];
else if(j == 0)
a[i][j] = a[i-1][j] + grid[i][j];
else
a[i][j] = min(a[i-1][j],a[i][j-1])+grid[i][j];
return a[w-1][h-1];
}
};

代码分析

很基础的动态规划题目。

到某点的距离最短，则在所有前一步里取得最短的再加上当前value。

即可以得到递推公式

f(i,j) = min(f(i-1,j),f(i,j-1))+grid(i,j)

题目中即为上方一格和左方一格中选取较小的。

注意当在第一行和第一列的时候要考虑边界情况