一、题目内容

二、题解

解法1：

思路

动态规划

• i=0,j=0
  • dp[i][j] = gift[i][j]
• i=0,j!=0
  • dp[i][j] = dp[i][j-1] + gift[i][j]
• i!=0,j=0
  • dp[i][j] = dp[i-1][j] + gift[i][j]
• i!=0,j!=0
  • dp[i][j] = Max{ dp[i][j-1] , dp[i-1][j] } + gift[i][j]

代码

class Solution {
    public int maxValue(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 && j == 0) {
                    continue;
                }
                if (i == 0) {
                    grid[i][j] += grid[i][j - 1];
                } else if (j == 0) {
                    grid[i][j] += grid[i - 1][j];
                } else {
                    // i!=0  j!=0
                    grid[i][j] = Math.max(grid[i - 1][j], grid[i][j - 1]) + grid[i][j];
                }
            }
        }
        return grid[m - 1][n - 1];
    }
}