# 1219. Path with Maximum Gold

yPhantom 2019年10月14日 83次浏览

In a gold mine `grid` of size `m * 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 `0` gold.
• 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:

• `1 <= grid.length, grid[i].length <= 15`
• `0 <= grid[i][j] <= 100`
• There are at most 25 cells containing gold.

## Solution

``````class Solution {
private final int[] d = {0, 1, 0, -1, 0};

public int getMaximumGold(int[][] grid) {
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid.length; j++) {
result = Math.max(result, dfs(grid, i, j, 0));
}
}
return result;
}

private int dfs(int[][] g, int i, int j, int sum) {
if (i < 0 || i >= g.length || j < 0 || j >= g.length || g[i][j] == 0 || g[i][j] > 100) {
return sum;
}
sum += g[i][j];
g[i][j] += 1000; // marked visited
int mx = 0;
for(int k = 0; k < 4; k++) {
mx = Math.max(mx, dfs(g, i + d[k], j + d[k + 1], sum));
}
g[i][j] -= 1000; // change back
return mx;
}
}
``````

DFS其实就是利用递归，通过限制条件来判断递归的终结，然后递归下去不断回溯上来。递归在伪代码的思想中可以理解为栈。BFS理解为队列。