ID | Title | Difficulty | |
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1091. Shortest Path in Binary Matrix
Medium
LeetCode
Problem
Given an n x n binary matrix grid, return the length of the shortest clear path in the matrix. If there is no clear path, return -1.
A clear path in a binary matrix is a path from the top-left cell (i.e., (0, 0)) to the bottom-right cell (i.e., (n - 1, n - 1)) such that:
- All the visited cells of the path are 0.
- All the adjacent cells of the path are 8-directionally connected (i.e., they are different and they share an edge or a corner).
The length of a clear path is the number of visited cells of this path.
Example 1:
Input: grid = [[0,1],[1,0]]
Output: 2
Example 2:
Input: grid = [[0,0,0],[1,1,0],[1,1,0]]
Output: 4
Example 3:
Input: grid = [[1,0,0],[1,1,0],[1,1,0]]
Output: -1
Code
class Solution {
int[][] dirs = new int[][] { { -1, -1 }, { -1, 0 }, { -1, 1 }, { 0, -1 }, { 0, 1 }, { 1, -1 }, { 1, 0 }, { 1, 1 } };
public int shortestPathBinaryMatrix(int[][] grid) {
int n = grid.length;
if (grid[0][0] == 1 || grid[n - 1][n - 1] == 1)
return -1;
Queue<int[]> queue = new LinkedList<>();
queue.offer(new int[] { 0, 0 });
grid[0][0] = 1;
int step = 1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int[] curr = queue.poll();
if (curr[0] == n - 1 && curr[1] == n - 1)
return step;
// 8 direction
for (int[] dir : dirs) {
int x = curr[0] + dir[0];
int y = curr[1] + dir[1];
if (x < 0 || x >= n || y < 0 || y >= n)
continue;
if (grid[x][y] == 1)
continue;
queue.offer(new int[] { x, y });
grid[x][y] = 1;
}
}
step++;
}
return -1;
}
}
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