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1463. 摘樱桃 II

English Version

题目描述

给你一个 rows x cols 的矩阵 grid 来表示一块樱桃地。 grid 中每个格子的数字表示你能获得的樱桃数目。

你有两个机器人帮你收集樱桃，机器人 1 从左上角格子 (0,0) 出发，机器人 2 从右上角格子 (0, cols-1) 出发。

请你按照如下规则，返回两个机器人能收集的最多樱桃数目：

从格子 (i,j) 出发，机器人可以移动到格子 (i+1, j-1)，(i+1, j) 或者 (i+1, j+1) 。
当一个机器人经过某个格子时，它会把该格子内所有的樱桃都摘走，然后这个位置会变成空格子，即没有樱桃的格子。
当两个机器人同时到达同一个格子时，它们中只有一个可以摘到樱桃。
两个机器人在任意时刻都不能移动到 grid 外面。
两个机器人最后都要到达 grid 最底下一行。

示例 1：

输入：grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
输出：24
解释：机器人 1 和机器人 2 的路径在上图中分别用绿色和蓝色表示。
机器人 1 摘的樱桃数目为 (3 + 2 + 5 + 2) = 12 。
机器人 2 摘的樱桃数目为 (1 + 5 + 5 + 1) = 12 。
樱桃总数为： 12 + 12 = 24 。

示例 2：

输入：grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
输出：28
解释：机器人 1 和机器人 2 的路径在上图中分别用绿色和蓝色表示。
机器人 1 摘的樱桃数目为 (1 + 9 + 5 + 2) = 17 。
机器人 2 摘的樱桃数目为 (1 + 3 + 4 + 3) = 11 。
樱桃总数为： 17 + 11 = 28 。

示例 3：

输入：grid = [[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
输出：22

示例 4：

输入：grid = [[1,1],[1,1]]
输出：4

提示：

rows == grid.length
cols == grid[i].length
2 <= rows, cols <= 70
0 <= grid[i][j] <= 100

解法

方法一：动态规划

我们定义 $f[i][j_1][j_2]$ 表示两个机器人分别在第 $i$ 行的位置 $j_1$ 和 $j_2$ 时能够摘到的最多樱桃数目。初始时 $f[0][0][n-1] = grid[0][0] + grid[0][n-1]$，其余值均为 $-1$。答案为 $\max_{0 \leq j_1, j_2 < n} f[m-1][j_1][j_2]$。

考虑 $f[i][j_1][j_2]$，如果 $j_1 \neq j_2$，那么机器人在第 $i$ 行能摘到的樱桃数目为 $grid[i][j_1] + grid[i][j_2]$；如果 $j_1 = j_2$，那么机器人在第 $i$ 行能摘到的樱桃数目为 $grid[i][j_1]$。我们可以枚举两个机器人的上一个状态 $f[i-1][y1][y2]$，其中 $y_1, y_2$ 分别是两个机器人在第 $i-1$ 行的位置，那么有 $y_1 \in {j_1-1, j_1, j_1+1}$ 且 $y_2 \in {j_2-1, j_2, j_2+1}$。状态转移方程如下：

$$ f[i][j_1][j_2] = \max_{y_1 \in {j_1-1, j_1, j_1+1}, y_2 \in {j_2-1, j_2, j_2+1}} f[i-1][y_1][y_2] + \begin{cases} grid[i][j_1] + grid[i][j_2], & j_1 \neq j_2 \ grid[i][j_1], & j_1 = j_2 \end{cases} $$

其中 $f[i-1][y_1][y_2]$ 为 $-1$ 时需要忽略。

最终的答案即为 $\max_{0 \leq j_1, j_2 < n} f[m-1][j_1][j_2]$。

时间复杂度 $O(m \times n^2)$，空间复杂度 $O(m \times n^2)$。其中 $m$ 和 $n$ 分别是网格的行数和列数。

Python3

class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        f = [[[-1] * n for _ in range(n)] for _ in range(m)]
        f[0][0][n - 1] = grid[0][0] + grid[0][n - 1]
        for i in range(1, m):
            for j1 in range(n):
                for j2 in range(n):
                    x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2])
                    for y1 in range(j1 - 1, j1 + 2):
                        for y2 in range(j2 - 1, j2 + 2):
                            if 0 <= y1 < n and 0 <= y2 < n and f[i - 1][y1][y2] != -1:
                                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x)
        return max(f[-1][j1][j2] for j1, j2 in product(range(n), range(n)))

Java

class Solution {
    public int cherryPickup(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][][] f = new int[m][n][n];
        for (var g : f) {
            for (var h : g) {
                Arrays.fill(h, -1);
            }
        }
        f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
        for (int i = 1; i < m; ++i) {
            for (int j1 = 0; j1 < n; ++j1) {
                for (int j2 = 0; j2 < n; ++j2) {
                    int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
                    for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                        for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                                f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x);
                            }
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int j1 = 0; j1 < n; ++j1) {
            for (int j2 = 0; j2 < n; ++j2) {
                ans = Math.max(ans, f[m - 1][j1][j2]);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int cherryPickup(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int f[m][n][n];
        memset(f, -1, sizeof(f));
        f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
        for (int i = 1; i < m; ++i) {
            for (int j1 = 0; j1 < n; ++j1) {
                for (int j2 = 0; j2 < n; ++j2) {
                    int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
                    for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                        for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x);
                            }
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int j1 = 0; j1 < n; ++j1) {
            for (int j2 = 0; j2 < n; ++j2) {
                ans = max(ans, f[m - 1][j1][j2]);
            }
        }
        return ans;
    }
};

Go

func cherryPickup(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	f := make([][][]int, m)
	for i := range f {
		f[i] = make([][]int, n)
		for j := range f[i] {
			f[i][j] = make([]int, n)
			for k := range f[i][j] {
				f[i][j][k] = -1
			}
		}
	}
	f[0][0][n-1] = grid[0][0] + grid[0][n-1]
	for i := 1; i < m; i++ {
		for j1 := 0; j1 < n; j1++ {
			for j2 := 0; j2 < n; j2++ {
				x := grid[i][j1]
				if j1 != j2 {
					x += grid[i][j2]
				}
				for y1 := j1 - 1; y1 <= j1+1; y1++ {
					for y2 := j2 - 1; y2 <= j2+1; y2++ {
						if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i-1][y1][y2] != -1 {
							f[i][j1][j2] = max(f[i][j1][j2], f[i-1][y1][y2]+x)
						}
					}
				}
			}
		}
	}
	for j1 := 0; j1 < n; j1++ {
		ans = max(ans, slices.Max(f[m-1][j1]))
	}
	return
}

TypeScript

function cherryPickup(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const f = Array.from({ length: m }, () =>
        Array.from({ length: n }, () => Array.from({ length: n }, () => -1)),
    );
    f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (let i = 1; i < m; ++i) {
        for (let j1 = 0; j1 < n; ++j1) {
            for (let j2 = 0; j2 < n; ++j2) {
                const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]);
                for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                    for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                        if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] !== -1) {
                            f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x);
                        }
                    }
                }
            }
        }
    }
    return Math.max(...f[m - 1].flat());
}

方法二：动态规划（空间优化）

注意到 $f[i][j_1][j_2]$ 的计算只和 $f[i-1][y_1][y_2]$ 有关，因此我们可以使用滚动数组优化空间复杂度，空间复杂度优化后的时间复杂度为 $O(n^2)$。

Python3

class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        f = [[-1] * n for _ in range(n)]
        g = [[-1] * n for _ in range(n)]
        f[0][n - 1] = grid[0][0] + grid[0][n - 1]
        for i in range(1, m):
            for j1 in range(n):
                for j2 in range(n):
                    x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2])
                    for y1 in range(j1 - 1, j1 + 2):
                        for y2 in range(j2 - 1, j2 + 2):
                            if 0 <= y1 < n and 0 <= y2 < n and f[y1][y2] != -1:
                                g[j1][j2] = max(g[j1][j2], f[y1][y2] + x)
            f, g = g, f
        return max(f[j1][j2] for j1, j2 in product(range(n), range(n)))

Java

class Solution {
    public int cherryPickup(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] f = new int[n][n];
        int[][] g = new int[n][n];
        for (int i = 0; i < n; ++i) {
            Arrays.fill(f[i], -1);
            Arrays.fill(g[i], -1);
        }
        f[0][n - 1] = grid[0][0] + grid[0][n - 1];
        for (int i = 1; i < m; ++i) {
            for (int j1 = 0; j1 < n; ++j1) {
                for (int j2 = 0; j2 < n; ++j2) {
                    int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
                    for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                        for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) {
                                g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x);
                            }
                        }
                    }
                }
            }
            int[][] t = f;
            f = g;
            g = t;
        }
        int ans = 0;
        for (int j1 = 0; j1 < n; ++j1) {
            for (int j2 = 0; j2 < n; ++j2) {
                ans = Math.max(ans, f[j1][j2]);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int cherryPickup(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int>> f(n, vector<int>(n, -1));
        vector<vector<int>> g(n, vector<int>(n, -1));
        f[0][n - 1] = grid[0][0] + grid[0][n - 1];
        for (int i = 1; i < m; ++i) {
            for (int j1 = 0; j1 < n; ++j1) {
                for (int j2 = 0; j2 < n; ++j2) {
                    int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
                    for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                        for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) {
                                g[j1][j2] = max(g[j1][j2], f[y1][y2] + x);
                            }
                        }
                    }
                }
            }
            swap(f, g);
        }
        int ans = 0;
        for (int j1 = 0; j1 < n; ++j1) {
            for (int j2 = 0; j2 < n; ++j2) {
                ans = max(ans, f[j1][j2]);
            }
        }
        return ans;
    }
};

Go

func cherryPickup(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	f := make([][]int, n)
	g := make([][]int, n)
	for i := range f {
		f[i] = make([]int, n)
		g[i] = make([]int, n)
		for j := range f[i] {
			f[i][j] = -1
			g[i][j] = -1
		}
	}
	f[0][n-1] = grid[0][0] + grid[0][n-1]
	for i := 1; i < m; i++ {
		for j1 := 0; j1 < n; j1++ {
			for j2 := 0; j2 < n; j2++ {
				x := grid[i][j1]
				if j1 != j2 {
					x += grid[i][j2]
				}
				for y1 := j1 - 1; y1 <= j1+1; y1++ {
					for y2 := j2 - 1; y2 <= j2+1; y2++ {
						if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1 {
							g[j1][j2] = max(g[j1][j2], f[y1][y2]+x)
						}
					}
				}
			}
		}
		f, g = g, f
	}
	for j1 := 0; j1 < n; j1++ {
		ans = max(ans, slices.Max(f[j1]))
	}
	return
}

TypeScript

function cherryPickup(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => -1));
    let g: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => -1));
    f[0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (let i = 1; i < m; ++i) {
        for (let j1 = 0; j1 < n; ++j1) {
            for (let j2 = 0; j2 < n; ++j2) {
                const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]);
                for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                    for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                        if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] !== -1) {
                            g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x);
                        }
                    }
                }
            }
        }
        [f, g] = [g, f];
    }
    return Math.max(...f.flat());
}

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README.md

1463. 摘樱桃 II

题目描述

解法

方法一：动态规划

Python3

Java

C++

Go

TypeScript

方法二：动态规划（空间优化）

Python3

Java

C++

Go

TypeScript

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1463. 摘樱桃 II

题目描述

解法

方法一：动态规划

Python3

Java

C++

Go

TypeScript

方法二：动态规划（空间优化）

Python3

Java

C++

Go

TypeScript