题解 | #礼物的最大价值#

#include <algorithm>
#include <sys/ucontext.h>
#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param grid int整型vector<vector<>> 
     * @return int整型
     */
    int maxValue(vector<vector<int> >& grid) {
        // write code here
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 0));  //含义到[i][j]的最大值
        dp[0][0] = grid[0][0];    //初始化首行和首列
        for (int i = 1; i < m ; i++){
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }
        for (int i = 1; i < n; i++){
            dp[0][i] = dp[0][i - 1] + grid[0][i];
        }
        for (int i = 1; i < m; i++) {       //遍历
            for (int j = 1; j < n; j++){
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];  //递推公式
            }
        }
        return dp[m - 1][n - 1];
    }
};