时间轴

2025-12-13

init


题目:

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:

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

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#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];
}
};