1、前言

2、思路

就是一个很普通的记忆化的 dfs，当然可以改成 dp

3、代码

class Solution {
     public int maxValue(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        int[][] memo = new int[grid.length][grid[0].length];
        return dfs(grid, 0, 0, memo);
    }

    private int dfs(int[][] grid, int i, int j, int[][] memo){
        if(i >= grid.length || j >= grid[0].length){
            return 0;
        }

        if(i == grid.length - 1 && j == grid[0].length - 1){
            return grid[grid.length - 1][grid[0].length - 1];
        }
        if(memo[i][j] != 0){
            return memo[i][j];
        }

        memo[i][j] = Math.max(dfs(grid, i + 1, j, memo), dfs(grid, i, j + 1, memo)) + grid[i][j];

        return memo[i][j];
    }
}

动态规划初始化，记得累加：

class Solution {
    public int maxValue(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        int m = grid.length, n = grid[0].length;
        int[][] dp = new int[m][n];
        dp[0][0] = grid[0][0];
        // 初始化的时候，记得是累加
        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++){
            for(int j = 1; j < n; j++){
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }

        return dp[m - 1][n - 1];
    }
}