题解 | JZ47 礼物的最大价值

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 n = grid.size();
        int m = grid[0].size();
        
        int dp[n][m];
        
        dp[0][0] = grid[0][0];
        for (int j = 1; j < m; j++) {
            dp[0][j] = dp[0][j-1] + grid[0][j];
        }
        for(int i = 1; i < n; i++) {
            dp[i][0] = dp[i-1][0] + grid[i][0];
        }
        
        
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < m; j++) {
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + grid[i][j];
            }
        }
        return dp[n-1][m-1];
    }
};

#
# 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整型二维数组 
# @return int整型
#
class Solution:
    def maxValue(self , grid: List[List[int]]) -> int:
        # write code here
        n = len(grid)
        m = len(grid[0])
        
        dp = [[0] * m for i in range(n)]

        dp[0][0] = grid[0][0]
        
        for j in range(1, m):
            dp[0][j] = dp[0][j-1] + grid[0][j]
        for i in range(1, n):
            dp[i][0] = dp[i-1][0] + grid[i][0]
            
        for i in range(1, n):
            for j in range(1, m):
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + grid[i][j]
        return dp[n-1][m-1]