64. Minimum Path Sum

Posted 2020-10-20 小预备

#week13

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.

 

分析：

这个动态规划的状态转换很明显：

sum[i][j]  = min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];

要么从上面走过来要么从左边走过来，才能够得到最小的，因为边的权值为正数，所以不可能是右边和下面

然后最后的结果就是sum[m - 1][n - 1]

 

题解：

 1 class Solution {
 2 public:
 3     int minPathSum(vector<vector<int>>& grid) {
 4         int m = grid.size();
 5         int n = grid[0].size(); 
 6         vector<vector<int> > sum(m, vector<int>(n, grid[0][0]));
 7         for (int i = 1; i < m; i++)
 8             sum[i][0] = sum[i - 1][0] + grid[i][0];
 9         for (int j = 1; j < n; j++)
10             sum[0][j] = sum[0][j - 1] + grid[0][j];
11         for (int i = 1; i < m; i++)
12             for (int j = 1; j < n; j++)
13                 sum[i][j]  = min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
14         return sum[m - 1][n - 1];
15     }
16 };