dp题,动归方程为:dp[i][j] = min(dp[i-1][j], dp[i][j-1])+grid[i][j],dp[i][j]表示到达坐标为(i,j)方格的最短路径。
class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(grid.empty()) return 0;
int row = grid.size(), col = grid[0].size();
vector<vector<int> > dp(row, vector<int>(col));
dp[0][0] = grid[0][0];
for(int i = 1; i < col; i++)
dp[0][i] = dp[0][i-1]+grid[0][i];
for(int i = 1; i < row; i++)
dp[i][0] = dp[i-1][0]+grid[i][0];
for(int i = 1; i < row; i++){
for(int j = 1; j < col; j++){
dp[i][j] = min(dp[i-1][j], dp[i][j-1])+grid[i][j];
}
}
return dp[row-1][col-1];
}
};