动态规划
给定一个包含非负整数的 m x n 网格,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小。
说明:每次只能向下或者向右移动一步。
输入:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
输出: 7
解释: 因为路径 1→3→1→1→1 的总和最小。
int minPathSum(vector<vector<int>>& grid) {
}
入门级动态规划
/
子问题: sub[i][j] = min(sub[i+1][j],sub[i][j+1]) + nums[i][j]
第一行:dp[i][j] = dp[i][j-1] + grid[i][j];
第一列:dp[i][j] = dp[i-1][j] + grid[i][j];
1 3 1
1 5 1
4 2 1
/
- 先把行和列初始化算好
```cpp
class Solution {
public:
int minPathSum(vector
>& grid) {
} };int row = grid.size();
int col = grid[0].size();
vector<vector<int>> dp(row, vector<int>(col, 0));
vectot<vector<int>> dp(row, vector<int>(col, 0));
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if(i ==0 && j == 0) dp[i][j] = grid[i][j];
else if(i == 0 && j != 0) dp[i][j] = dp[i][j-1] + grid[i][j];
else if(j == 0 && i != 0) dp[i][j] = dp[i-1][j] + grid[i][j];
else dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j];
}
}
return dp[row-1][col-1];
```