1391. Check if There is a Valid Path in a Grid

Description

Given a m x n grid. Each cell of the grid represents a street. The street of grid[i][j] can be:

  • 1 which means a street connecting the left cell and the right cell.
  • 2 which means a street connecting the upper cell and the lower cell.
  • 3 which means a street connecting the left cell and the lower cell.
  • 4 which means a street connecting the right cell and the lower cell.
  • 5 which means a street connecting the left cell and the upper cell.
  • 6 which means a street connecting the right cell and the upper cell.
img

You will initially start at the street of the upper-left cell (0,0). A valid path in the grid is a path which starts from the upper left cell (0,0) and ends at the bottom-right cell (m - 1, n - 1). The path should only follow the streets.

Notice that you are not allowed to change any street.

Return true if there is a valid path in the grid or false otherwise.

Example 1:

img
Input: grid = [[2,4,3],[6,5,2]]
Output: true
Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).

Example 2:

img
Input: grid = [[1,2,1],[1,2,1]]
Output: false
Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)

Example 3:

Input: grid = [[1,1,2]]
Output: false
Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).

Example 4:

Input: grid = [[1,1,1,1,1,1,3]]
Output: true

Example 5:

Input: grid = [[2],[2],[2],[2],[2],[2],[6]]
Output: true

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 300
  • 1 <= grid[i][j] <= 6

Solution

这题目翻译过来就是:判断点(0, 0)是否和点(m – 1, n – 1)连通。连通性问题我们直接采用并查集的方法。

这道题属于图切割的并查集。

根据题意给出的6种前进方式,我们将原来的矩阵按如下方式切割。

grid[0][0]点仍然是grid[0][0]

grid[i][j]点就变成了grid[2i][2j]

此时对于点(i, j)其上下左右4个边恰好和题中给出的6个方向契合。

Union条件

if grid[i][j] in [2, 5, 6]: connect center and top
if grid[i][j] in [1, 3, 5]: connect center and left
if grid[i][j] in [2, 3, 4]: connect center and bottom
if grid[i][j] in [1, 4, 6]: connect center and right

结果

uf[0][0]是否与uf[2*(m-1), 2*(n-1)]相连

时间复杂度

Time O(MN) * O(UF)
Space O(MN)

Python

class Solution(object):
    def hasValidPath(self, grid):
        m, n = len(grid), len(grid[0])
        uf = {(i, j) : (i, j) for j in xrange(-1, 2*m + 1) for i in xrange(-1, 2*n + 1)}
        
        def find(x):
            while x != uf[x]:
                uf[x] = uf[uf[x]]
                x = uf[x]
            return x
        
        def union(x, y, dx, dy):
            root = find((x, y))
            droot = find((x + dx, y + dy))
            if root != droot:
                uf[root] = droot
        
        for i in xrange(m):
            for j in xrange(n):
                if grid[i][j] in (2, 5, 6): union(2*i, 2*j, -1, 0)
                if grid[i][j] in (2, 3, 4): union(2*i, 2*j, 1, 0)
                if grid[i][j] in (1, 3, 5): union(2*i, 2*j, 0, -1)
                if grid[i][j] in (1, 4, 6): union(2*i, 2*j, 0, 1)
        
        return find((0, 0)) == find((2*m - 2, 2*n - 2))

Java

class Solution {
    public boolean hasValidPath(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int t = 2*n + 1;
        UF uf = new UF((2*m + 1)*(2*n + 1));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1 || grid[i][j] == 3 || grid[i][j] == 5) {
                    uf.union(g(2*i + 1, 2*j + 1, t), g(2*i + 1, 2*j, t));
                }
                if (grid[i][j] == 1 || grid[i][j] == 4 || grid[i][j] == 6) {
                    uf.union(g(2*i + 1, 2*j + 1, t), g(2*i + 1, 2*j + 2, t));
                }
                if (grid[i][j] == 2 || grid[i][j] == 3 || grid[i][j] == 4) {
                    uf.union(g(2*i + 1, 2*j + 1, t), g(2*i + 2, 2*j + 1, t));
                }
                if (grid[i][j] == 2 || grid[i][j] == 5 || grid[i][j] == 6) {
                    uf.union(g(2*i + 1, 2*j + 1, t), g(2*i, 2*j + 1, t));
                }
            }
        }
        return uf.isConnected(g(1, 1, t), g(2*m - 1, 2*n - 1, t));
    }
    
    public int g(int r, int c, int n) {
        return r*n + c;
    }
    
    class UF {
        private int[] parents;
        
        public UF(int n) {
            parents = new int[n];
            
            for (int i = 0; i < n; i++) {
                parents[i] = i;
            }
        }
        
        public int find(int node) {
            while (node != parents[node]) {
                parents[node] = parents[parents[node]];
                node = parents[node];
            }
            return node;
        }
        
        public void union(int node1, int node2) {
            int root1 = find(node1);
            int root2 = find(node2);
            if (root1 != root2) {
                parents[root1] = root2;
            }
        }
        
        public boolean isConnected(int node1, int node2) {
            return find(node1) == find(node2);
        }
    }
}

暂无评论

发表评论

电子邮件地址不会被公开。 必填项已用*标注

浙ICP备19002997号