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困难
数组
动态规划
矩阵

English Version

题目描述

给你一个 n x n 的网格 grid ,代表一块樱桃地,每个格子由以下三种数字的一种来表示:

  • 0 表示这个格子是空的,所以你可以穿过它。
  • 1 表示这个格子里装着一个樱桃,你可以摘到樱桃然后穿过它。
  • -1 表示这个格子里有荆棘,挡着你的路。

请你统计并返回:在遵守下列规则的情况下,能摘到的最多樱桃数:

  • 从位置 (0, 0) 出发,最后到达 (n - 1, n - 1) ,只能向下或向右走,并且只能穿越有效的格子(即只可以穿过值为 0 或者 1 的格子);
  • 当到达 (n - 1, n - 1) 后,你要继续走,直到返回到 (0, 0) ,只能向上或向左走,并且只能穿越有效的格子;
  • 当你经过一个格子且这个格子包含一个樱桃时,你将摘到樱桃并且这个格子会变成空的(值变为 0 );
  • 如果在 (0, 0)(n - 1, n - 1) 之间不存在一条可经过的路径,则无法摘到任何一个樱桃。

 

示例 1:

输入:grid = [[0,1,-1],[1,0,-1],[1,1,1]]
输出:5
解释:玩家从 (0, 0) 出发:向下、向下、向右、向右移动至 (2, 2) 。
在这一次行程中捡到 4 个樱桃,矩阵变成 [[0,1,-1],[0,0,-1],[0,0,0]] 。
然后,玩家向左、向上、向上、向左返回起点,再捡到 1 个樱桃。
总共捡到 5 个樱桃,这是最大可能值。

示例 2:

输入:grid = [[1,1,-1],[1,-1,1],[-1,1,1]]
输出:0

 

提示:

  • n == grid.length
  • n == grid[i].length
  • 1 <= n <= 50
  • grid[i][j] 为 -10 或 1
  • grid[0][0] != -1
  • grid[n - 1][n - 1] != -1

解法

方法一:动态规划

根据题目描述,玩家从 $(0, 0)$ 出发,到达 $(n-1, n-1)$ 后,又重新返回到起始点 $(0, 0)$,我们可以视为玩家两次从 $(0, 0)$ 出发,到达 $(n-1, n-1)$

因此,我们定义 $f[k][i_1][i_2]$ 表示两次都走了 $k$ 步,分别到达 $(i_1, k-i_1)$$(i_2, k-i_2)$ 时,能够摘到的最多樱桃数。初始时 $f[0][0][0] = grid[0][0]$。其余 $f[k][i_1][i_2]$ 的初始值为负无穷。答案为 $\max(0, f[2n-2][n-1][n-1])$

我们可以根据题目描述,得到状态转移方程:

$$ f[k][i_1][i_2] = \max(f[k-1][x_1][x_2] + t, f[k][i_1][i_2]) $$

其中 $t$ 表示 $(i_1, k-i_1)$$(i_2, k-i_2)$ 位置上的樱桃数,而 $x_1, x_2$ 分别表示 $(i_1, k-i_1)$$(i_2, k-i_2)$ 的前一步位置。

时间复杂度 $O(n^3)$,空间复杂度 $O(n^3)$。其中 $n$ 表示网格的边长。

Python3

class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        n = len(grid)
        f = [[[-inf] * n for _ in range(n)] for _ in range((n << 1) - 1)]
        f[0][0][0] = grid[0][0]
        for k in range(1, (n << 1) - 1):
            for i1 in range(n):
                for i2 in range(n):
                    j1, j2 = k - i1, k - i2
                    if (
                        not 0 <= j1 < n
                        or not 0 <= j2 < n
                        or grid[i1][j1] == -1
                        or grid[i2][j2] == -1
                    ):
                        continue
                    t = grid[i1][j1]
                    if i1 != i2:
                        t += grid[i2][j2]
                    for x1 in range(i1 - 1, i1 + 1):
                        for x2 in range(i2 - 1, i2 + 1):
                            if x1 >= 0 and x2 >= 0:
                                f[k][i1][i2] = max(f[k][i1][i2], f[k - 1][x1][x2] + t)
        return max(0, f[-1][-1][-1])

Java

class Solution {
    public int cherryPickup(int[][] grid) {
        int n = grid.length;
        int[][][] f = new int[n * 2][n][n];
        f[0][0][0] = grid[0][0];
        for (int k = 1; k < n * 2 - 1; ++k) {
            for (int i1 = 0; i1 < n; ++i1) {
                for (int i2 = 0; i2 < n; ++i2) {
                    int j1 = k - i1, j2 = k - i2;
                    f[k][i1][i2] = Integer.MIN_VALUE;
                    if (j1 < 0 || j1 >= n || j2 < 0 || j2 >= n || grid[i1][j1] == -1
                        || grid[i2][j2] == -1) {
                        continue;
                    }
                    int t = grid[i1][j1];
                    if (i1 != i2) {
                        t += grid[i2][j2];
                    }
                    for (int x1 = i1 - 1; x1 <= i1; ++x1) {
                        for (int x2 = i2 - 1; x2 <= i2; ++x2) {
                            if (x1 >= 0 && x2 >= 0) {
                                f[k][i1][i2] = Math.max(f[k][i1][i2], f[k - 1][x1][x2] + t);
                            }
                        }
                    }
                }
            }
        }
        return Math.max(0, f[n * 2 - 2][n - 1][n - 1]);
    }
}

C++

class Solution {
public:
    int cherryPickup(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<vector<int>>> f(n << 1, vector<vector<int>>(n, vector<int>(n, -1e9)));
        f[0][0][0] = grid[0][0];
        for (int k = 1; k < n * 2 - 1; ++k) {
            for (int i1 = 0; i1 < n; ++i1) {
                for (int i2 = 0; i2 < n; ++i2) {
                    int j1 = k - i1, j2 = k - i2;
                    if (j1 < 0 || j1 >= n || j2 < 0 || j2 >= n || grid[i1][j1] == -1 || grid[i2][j2] == -1) {
                        continue;
                    }
                    int t = grid[i1][j1];
                    if (i1 != i2) {
                        t += grid[i2][j2];
                    }
                    for (int x1 = i1 - 1; x1 <= i1; ++x1) {
                        for (int x2 = i2 - 1; x2 <= i2; ++x2) {
                            if (x1 >= 0 && x2 >= 0) {
                                f[k][i1][i2] = max(f[k][i1][i2], f[k - 1][x1][x2] + t);
                            }
                        }
                    }
                }
            }
        }
        return max(0, f[n * 2 - 2][n - 1][n - 1]);
    }
};

Go

func cherryPickup(grid [][]int) int {
	n := len(grid)
	f := make([][][]int, (n<<1)-1)
	for i := range f {
		f[i] = make([][]int, n)
		for j := range f[i] {
			f[i][j] = make([]int, n)
		}
	}
	f[0][0][0] = grid[0][0]
	for k := 1; k < (n<<1)-1; k++ {
		for i1 := 0; i1 < n; i1++ {
			for i2 := 0; i2 < n; i2++ {
				f[k][i1][i2] = int(-1e9)
				j1, j2 := k-i1, k-i2
				if j1 < 0 || j1 >= n || j2 < 0 || j2 >= n || grid[i1][j1] == -1 || grid[i2][j2] == -1 {
					continue
				}
				t := grid[i1][j1]
				if i1 != i2 {
					t += grid[i2][j2]
				}
				for x1 := i1 - 1; x1 <= i1; x1++ {
					for x2 := i2 - 1; x2 <= i2; x2++ {
						if x1 >= 0 && x2 >= 0 {
							f[k][i1][i2] = max(f[k][i1][i2], f[k-1][x1][x2]+t)
						}
					}
				}
			}
		}
	}
	return max(0, f[n*2-2][n-1][n-1])
}

TypeScript

function cherryPickup(grid: number[][]): number {
    const n: number = grid.length;
    const f: number[][][] = Array.from({ length: n * 2 - 1 }, () =>
        Array.from({ length: n }, () => Array.from({ length: n }, () => -Infinity)),
    );
    f[0][0][0] = grid[0][0];
    for (let k = 1; k < n * 2 - 1; ++k) {
        for (let i1 = 0; i1 < n; ++i1) {
            for (let i2 = 0; i2 < n; ++i2) {
                const [j1, j2]: [number, number] = [k - i1, k - i2];
                if (
                    j1 < 0 ||
                    j1 >= n ||
                    j2 < 0 ||
                    j2 >= n ||
                    grid[i1][j1] == -1 ||
                    grid[i2][j2] == -1
                ) {
                    continue;
                }
                const t: number = grid[i1][j1] + (i1 != i2 ? grid[i2][j2] : 0);
                for (let x1 = i1 - 1; x1 <= i1; ++x1) {
                    for (let x2 = i2 - 1; x2 <= i2; ++x2) {
                        if (x1 >= 0 && x2 >= 0) {
                            f[k][i1][i2] = Math.max(f[k][i1][i2], f[k - 1][x1][x2] + t);
                        }
                    }
                }
            }
        }
    }
    return Math.max(0, f[n * 2 - 2][n - 1][n - 1]);
}

JavaScript

/**
 * @param {number[][]} grid
 * @return {number}
 */
var cherryPickup = function (grid) {
    const n = grid.length;
    const f = Array.from({ length: n * 2 - 1 }, () =>
        Array.from({ length: n }, () => Array.from({ length: n }, () => -Infinity)),
    );
    f[0][0][0] = grid[0][0];
    for (let k = 1; k < n * 2 - 1; ++k) {
        for (let i1 = 0; i1 < n; ++i1) {
            for (let i2 = 0; i2 < n; ++i2) {
                const [j1, j2] = [k - i1, k - i2];
                if (
                    j1 < 0 ||
                    j1 >= n ||
                    j2 < 0 ||
                    j2 >= n ||
                    grid[i1][j1] == -1 ||
                    grid[i2][j2] == -1
                ) {
                    continue;
                }
                const t = grid[i1][j1] + (i1 != i2 ? grid[i2][j2] : 0);
                for (let x1 = i1 - 1; x1 <= i1; ++x1) {
                    for (let x2 = i2 - 1; x2 <= i2; ++x2) {
                        if (x1 >= 0 && x2 >= 0) {
                            f[k][i1][i2] = Math.max(f[k][i1][i2], f[k - 1][x1][x2] + t);
                        }
                    }
                }
            }
        }
    }
    return Math.max(0, f[n * 2 - 2][n - 1][n - 1]);
};