描述
You are given two integers m and n representing a 0-indexed m x n grid. You are also given two 2D integer arrays guards and walls where guards[i] = [rowi, coli] and walls[j] = [rowj, colj] represent the positions of the ith guard and jth wall respectively. A guard can see every cell in the four cardinal directions (north, east, south, or west) starting from their position unless obstructed by a wall or another guard. A cell is guarded if there is at least one guard that can see it. Return the number of unoccupied cells that are not guarded.
Example 1:
Input: m = 4, n = 6, guards = [[0,0],[1,1],[2,3]], walls = [[0,1],[2,2],[1,4]]
Output: 7
Explanation: The guarded and unguarded cells are shown in red and green respectively in the above diagram.
There are a total of 7 unguarded cells, so we return 7.
Example 2:
Input: m = 3, n = 3, guards = [[1,1]], walls = [[0,1],[1,0],[2,1],[1,2]]
Output: 4
Explanation: The unguarded cells are shown in green in the above diagram.
There are a total of 4 unguarded cells, so we return 4.
Note:
1 <= m, n <= 10^5
2 <= m * n <= 10^5
1 <= guards.length, walls.length <= 5 * 10^4
2 <= guards.length + walls.length <= m * n
guards[i].length == walls[j].length == 2
0 <= rowi, rowj < m
0 <= coli, colj < n
All the positions in guards and walls are unique.
解析
根据题意，给定两个整数 m 和 n ，表示一个 0 索引的 m x n 网格。 给定两个二维整数数组 guards 和 walls ，其中 guards[i] = [rowi, coli] 和 walls[j] = [rowj, colj] 分别代表第 i 个保安和第 j 个墙的位置。
守卫可以从他们的位置开始看到四个主要方向（北、东、南或西）的每个单元，除非被墙或其他守卫阻挡。 返回未被看守的单元格的数量。
这道题其实就是一个遍历格子的问题，思路也比较简单，初始化一个全都是 0 的 grid ，然后将保安和墙的格子 2 表示这些位置已经有了站位或者阻挡，然后再遍历每个保安能看到的四个方向的格子，如果 grid[i][j] 上的数字为 0 或者 1 表示其还没有保卫或者已经保卫，然后将 grid[i][j] 设置为 1 即可，遍历结束只需要返回 grid 中还是 0 的格子数量即可。
时间复杂度为 O(m*n) ，空间复杂度为 O(m*n) 。
解答
class Solution(object):
def countUnguarded(self, m, n, guards, walls):
"""
:type m: int
:type n: int
:type guards: List[List[int]]
:type walls: List[List[int]]
:rtype: int
"""
grid = [[0]*n for _ in range(m)]
for i,j in guards:
grid[i][j] = 2
for i,j in walls:
grid[i][j] = 2
for i,j in guards:
for dx,dy in [[-1,0],[1,0],[0,-1],[0,1]]:
x = i+dx
y = j+dy
while 0<=x<m and 0<=y<n and (grid[x][y]==0 or grid[x][y]==1):
grid[x][y] = 1
x += dx
y += dy
return sum(grid[i][j]==0 for i in range(m) for j in range(n))
运行结果
47 / 47 test cases passed.
Status: Accepted
Runtime: 2600 ms
Memory Usage: 45.1 MB
原题链接
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