1091. Shortest Path in Binary Matrix (M)

Shortest Path in Binary Matrix (M)

题目

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

• Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
• C_1 is at location (0, 0) (ie. has value grid[0][0])
• C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
• If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right. If such a path does not exist, return -1.

Example 1:

Input: [[0,1],[1,0]]

Output: 2

Example 2:

Input: [[0,0,0],[1,1,0],[1,1,0]]

Output: 4

Note:

1. 1 <= grid.length == grid[0].length <= 100
2. grid[r][c] is 0 or 1

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题意

给定一个矩阵，其中值为1的格子不能走，为0的格子可以走。从每个格子出发有8个方向可走。求从左上到右下的最短路径。

思路

每条边的权值都为1，直接用BFS搜索最短路。

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代码实现

Java

class Solution {
    public int shortestPathBinaryMatrix(int[][] grid) {
        if (grid[0][0] == 1) return -1;
        
        int steps = 0;
        int n = grid.length;
        Queue<int[]> q = new ArrayDeque<>();
        boolean[][] visited = new boolean[n][n];
        q.offer(new int[]{0, 0});
        visited[0][0] = true;
        while (!q.isEmpty()) {
            int size = q.size();
            steps++;
            while (size > 0) {
                int[] cur = q.poll();
                int i = cur[0], j = cur[1];
                if (i == n - 1 && j == n - 1) {
                    return steps;
                }
                for (int x = -1; x <= 1; x++) {
                    for (int y = -1; y <= 1; y++) {
                        if (x != 0 || y != 0) {
                            int nextI = i + x, nextJ = j + y;
                            if (isValid(nextI, nextJ, n) && !visited[nextI][nextJ] && grid[nextI][nextJ] == 0) {
                                visited[nextI][nextJ] = true;
                                q.offer(new int[]{nextI, nextJ});
                            }
                        }
                    }
                }
                size--;
            }
        }
        return -1;
    }

    private boolean isValid(int i, int j,int n){
        return i < n && i >= 0 && j < n && j >= 0;
    }
}