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

这道题是方块上下左右都可以走,其实题意里面已经表述的很明确了,这道题应该用深度优先遍历,还是经验不足吧。

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[0].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[0].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。

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