分析

• 本题的考点：动态规划。

• 状态表示f[i][j]：从起点到达坐标点(i, j)的所有路径中权值和的最小值。

• 状态转移：f[i][j] = min(f[i-1][j], f[i][j-1]) + grid[i][j]。

代码

• C++

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {

        int n = grid.size(), m = grid[0].size();
        vector<vector<int>> f(n, vector<int>(m, INT_MAX));
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (!i && !j) f[i][j] = grid[i][j];
                else {
                    if (i) f[i][j] = min(f[i][j], f[i - 1][j] + grid[i][j]);
                    if (j) f[i][j] = min(f[i][j], f[i][j - 1] + grid[i][j]);
                }
        return f[n - 1][m - 1];
    }
};

• Java

class Solution {
    public int minPathSum(int[][] grid) {

        int n = grid.length, m = grid[0].length;
        int[][] f = new int[n][m];
        for (int i = 0; i < n; i++) Arrays.fill(f[i], Integer.MAX_VALUE);
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (i == 0 && j == 0) f[i][j] = grid[i][j];
                else {
                    if (i != 0) f[i][j] = Math.min(f[i][j], f[i - 1][j] + grid[i][j]);
                    if (j != 0) f[i][j] = Math.min(f[i][j], f[i][j - 1] + grid[i][j]);
                }
        return f[n - 1][m - 1];
    }
}

时空复杂度分析

• 时间复杂度：$O(n \times m)$，m、n为行数、列数。

• 空间复杂度：$O(n \times m)$。