LeetCode 1219. 黄金矿工

1219. 黄金矿工

Solution

思路：第一个想法是有点类似数塔dp的感觉，但是这里的起点是随机，而且规模不大，所以可以逐个枚举起点，然后dfs搜索即可。长时间不写，判断新点是否可行时，传入了旧的点，人麻了。

class Solution {
    int ans = 0;
    int[][] grids;
    int[][] dir = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
    int m, n;
    public int getMaximumGold(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        grids = grid;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] != 0) {
                    dfs(i, j, 0);
                }
            }
        }
        return ans;
    }
    void dfs(int x, int y, int gold) {
        int res = grids[x][y];
        ans = Math.max(ans, res + gold);
        grids[x][y] = 0;
        for (int i = 0; i < 4; i++) {
            int newX = x + dir[i][0];
            int newY = y + dir[i][1];
            if (check(newX, newY)) {
                dfs(newX, newY, res + gold);
            }
        }
        grids[x][y] = res;
    }
    boolean check(int x, int y) {
        if (x < 0 || x >= m || y < 0 || y >= n) return false;
        if (grids[x][y] == 0) return false;
        return true;
    }
}

标签：grids,int,res,1219,dfs,grid,ans,黄金矿工,LeetCode
来源： https://www.cnblogs.com/ACMerszl/p/15864535.html