In a gold mine grid of size m x n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.
Return the maximum amount of gold you can collect under the conditions:
- Every time you are located in a cell you will collect all the gold in that cell.
- From your position, you can walk one step to the left, right, up, or down.
- You can't visit the same cell more than once.
- Never visit a cell with
0gold. - You can start and stop collecting gold from any position in the grid that has some gold.
Example 1:
Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
Output: 24
Explanation:
[[0,6,0],
[5,8,7],
[0,9,0]]
Path to get the maximum gold, 9 -> 8 -> 7.
Example 2:
Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
Output: 28
Explanation:
[[1,0,7],
[2,0,6],
[3,4,5],
[0,3,0],
[9,0,20]]
Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 150 <= grid[i][j] <= 100- There are at most 25 cells containing gold.
这道题给了一个 m by n 的二维数组 grid，说是里面的数字代表金子的数量，0表示没有金子。现在可以选一个任意的起点，可以朝四个方向走，条件的是不能越界，不能走重复的位置，以及不能走值为0的地方，现在问最多能获得多少的金子。这道题虽然说也是一道迷宫遍历的问题，也是求极值的问题，但并不是求最短步数，而是求路径值之和最大，那么显然 BFS 就不太适合了，因为这里严格限制了不能走重复路径，并且要统计每一条路径之和，用 DFS 是坠好的。这道题的时间卡的非常的严格，博主最开始写的一个版本，由于用了 HashSet 来记录访问过的位置，都超时了，于是只能用 grid 数组本身来记录，首先遍历所有的位置，跳过所有为0的位置，对于有金子的位置，调用递归函数。
为了节省时间，这里的递归函数都加上了返回值，博主一般是不喜欢加返回值的。在递归函数中，首先判断当前位置是否越界，且是否有金子，不满足的话直接返回0。然后此时记录当前位置的金子数到一个变量 val 中，然后将当前位置的值置为0，表示访问过了，然后对其四个邻居位置调用递归函数，将最大值取出来放到变量 mx 中，之后将当前位置恢复为 val 值，并返回 mx+val 即可。用所有非0位置为起点调用递归函数的返回值中取最大值就是所求结果，参见代码如下：
class Solution {
public:
int getMaximumGold(vector<vector<int>>& grid) {
int res = 0, m = grid.size(), n = grid[0].size();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 0) continue;
res = max(res, helper(grid, i, j));
}
}
return res;
}
int helper(vector<vector<int>>& grid, int i, int j) {
int m = grid.size(), n = grid[0].size();
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) return 0;
int val = grid[i][j], mx = 0;
grid[i][j] = 0;
mx = max({helper(grid, i + 1, j), helper(grid, i - 1, j), helper(grid, i, j + 1), helper(grid, i, j - 1)});
grid[i][j] = val;
return mx + val;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1219
参考资料：
https://leetcode.com/problems/path-with-maximum-gold/
