面试经典150题 P64 最小路径和

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时间轴

2025-12-13

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题目：

P64 最小路径和

https://leetcode.cn/problems/minimum-path-sum/description/?envType=study-plan-v2&envId=top-interview-150

由于只能向右或向下，那某一点的上一步只能是左或上，设置 dp[i][j]表示到达 grid[i][j]的最小路径长度。有：

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

#include <vector>
using std::vector;

class Solution {
    public:
        int minPathSum(vector<vector<int> > &grid)
        {
                // dp[i][j]表示到达grid[i][j]的最小路径长度
                int i, j, m = grid.size(), n = grid[0].size();
                vector<vector<int> > dp(m, vector<int>(n, 0));

                dp[0][0] = grid[0][0];

                for (j = 1; j < n; j++) {
                        dp[0][j] = dp[0][j - 1] + grid[0][j];
                }
                for (i = 1; i < m; i++) {
                        dp[i][0] = dp[i - 1][0] + grid[i][0];
                }

                for (i = 1; i < m; i++) {
                        for (j = 1; j < n; j++) {
                                dp[i][j] = std::min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
                        }
                }
                return dp[m - 1][n - 1];
        }
};

leetcode hot 100 rewrite

#include <vector>
using std::vector;

class Solution {
    public:
        int minPathSum(vector<vector<int> > &grid)
        {
                int i, j, m = grid.size(), n = grid[0].size();
                // m == grid.length
                // n == grid[i].length
                // 1 <= m, n <= 200
                // 0 <= grid[i][j] <= 200

                // 每次只能向下或者向右移动一步。
                vector<vector<int> > dp(m, vector<int>(n, 0));

                dp[0][0] = grid[0][0];
                for (i = 1; i < m; i++)
                        dp[i][0] = dp[i - 1][0] + grid[i][0];

                for (j = 1; j < n; j++)
                        dp[0][j] = dp[0][j - 1] + grid[0][j];

                for (i = 1; i < m; i++) {
                        for (j = 1; j < n; j++)
                                dp[i][j] = std::min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
                }

                return dp[m - 1][n - 1];
        }
};