【leetcode】 64. Minimum Path Sum 动态规划

最新推荐文章于 2020-05-19 10:31:11 发布

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最新推荐文章于 2020-05-19 10:31:11 发布

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分类专栏： leetcode 【leetcode】题解

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本文介绍了一个经典算法问题——寻找矩阵中从左上角到右下角的最小路径和，只能向右或向下移动。通过动态规划的方法给出了详细的解决方案，并提供了完整的C++代码实现。

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:

Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

题解:入门题注意边界

    class Solution {
    public:
        int minPathSum(vector<vector<int>>& grid) {
            int n = grid.size();
            int m = grid[0].size();
            int dp[500][500];
            memset(dp, 0x3f, sizeof(dp));
            dp[0][0] = grid[0][0];
            for(int i = 1;i < n; i++){
                dp[i][0] = min(dp[i][0], dp[i-1][0] + grid[i][0]);
            }
            for(int i = 1;i < m; i++){
                dp[0][i] = min(dp[0][i], dp[0][i-1] + grid[0][i]);
            }
            for(int i = 1;i < n; i++){
                for(int j = 1;j < m; j++){
                    dp[i][j] = min(dp[i-1][j]+grid[i][j], dp[i][j]);
                    dp[i][j] = min(dp[i][j-1]+grid[i][j], dp[i][j]);
                }
            }
            return dp[n-1][m-1];
        }
    };