
题目描述
给你两个整数 m 和 n 表示一个下标从 0 开始的 m x n 网格图。同时给你两个二维整数数组 guards 和 walls ,其中 guards[i] = [rowi, coli] 且 walls[j] = [rowj, colj] ,分别表示第 i 个警卫和第 j 座墙所在的位置。
一个警卫能看到 4 个坐标轴方向(即东、南、西、北)的 所有 格子,除非他们被一座墙或者另外一个警卫 挡住 了视线。如果一个格子能被 至少 一个警卫看到,那么我们说这个格子被 保卫 了。
请你返回空格子中,有多少个格子是 没被保卫 的。
示例 1:

输入:m = 4, n = 6, guards = [[0,0],[1,1],[2,3]], walls = [[0,1],[2,2],[1,4]]
输出:7
解释:上图中,被保卫和没有被保卫的格子分别用红色和绿色表示。
总共有 7 个没有被保卫的格子,所以我们返回 7 。
示例 2:

输入:m = 3, n = 3, guards = [[1,1]], walls = [[0,1],[1,0],[2,1],[1,2]]
输出:4
解释:上图中,没有被保卫的格子用绿色表示。
总共有 4 个没有被保卫的格子,所以我们返回 4 。
提示:
1 <= m, n <= 105 2 <= m * n <= 105 1 <= guards.length, walls.length <= 5 * 104 2 <= guards.length + walls.length <= m * n guards[i].length == walls[j].length == 2 0 <= rowi, rowj < m 0 <= coli, colj < n guards 和 walls 中所有位置 互不相同 。
解法
方法一:模拟
我们创建一个 \(m \times n\) 的二维数组 \(g\),其中 \(g[i][j]\) 表示第 \(i\) 行第 \(j\) 列的格子。初始时 \(g[i][j]\) 的值为 \(0\),表示该格子没有被保卫。
然后遍历所有的警卫和墙,将 \(g[i][j]\) 的值置为 \(2\),这些位置不能被访问。
接下来,我们遍历所有警卫的位置,从该位置出发,向四个方向进行模拟,直到遇到墙或警卫,或者越界。在模拟的过程中,将遇到的格子的值置为 \(1\),表示该格子被保卫。
最后,我们遍历 \(g\),统计值为 \(0\) 的格子的个数,即为答案。
时间复杂度 \(O(m \times n)\),空间复杂度 \(O(m \times n)\)。其中 \(m\) 和 \(n\) 分别为网格的行数和列数。
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17 | class Solution:
def countUnguarded(
self, m: int, n: int, guards: List[List[int]], walls: List[List[int]]
) -> int:
g = [[0] * n for _ in range(m)]
for i, j in guards:
g[i][j] = 2
for i, j in walls:
g[i][j] = 2
dirs = (-1, 0, 1, 0, -1)
for i, j in guards:
for a, b in pairwise(dirs):
x, y = i, j
while 0 <= x + a < m and 0 <= y + b < n and g[x + a][y + b] < 2:
x, y = x + a, y + b
g[x][y] = 1
return sum(v == 0 for row in g for v in row)
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32 | class Solution {
public int countUnguarded(int m, int n, int[][] guards, int[][] walls) {
int[][] g = new int[m][n];
for (var e : guards) {
g[e[0]][e[1]] = 2;
}
for (var e : walls) {
g[e[0]][e[1]] = 2;
}
int[] dirs = {-1, 0, 1, 0, -1};
for (var e : guards) {
for (int k = 0; k < 4; ++k) {
int x = e[0], y = e[1];
int a = dirs[k], b = dirs[k + 1];
while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
x += a;
y += b;
g[x][y] = 1;
}
}
}
int ans = 0;
for (var row : g) {
for (int v : row) {
if (v == 0) {
++ans;
}
}
}
return ans;
}
}
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30 | class Solution {
public:
int countUnguarded(int m, int n, vector<vector<int>>& guards, vector<vector<int>>& walls) {
int g[m][n];
memset(g, 0, sizeof(g));
for (auto& e : guards) {
g[e[0]][e[1]] = 2;
}
for (auto& e : walls) {
g[e[0]][e[1]] = 2;
}
int dirs[5] = {-1, 0, 1, 0, -1};
for (auto& e : guards) {
for (int k = 0; k < 4; ++k) {
int x = e[0], y = e[1];
int a = dirs[k], b = dirs[k + 1];
while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
x += a;
y += b;
g[x][y] = 1;
}
}
}
int ans = 0;
for (auto& row : g) {
ans += count(row, row + n, 0);
}
return ans;
}
};
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31 | func countUnguarded(m int, n int, guards [][]int, walls [][]int) (ans int) {
g := make([][]int, m)
for i := range g {
g[i] = make([]int, n)
}
for _, e := range guards {
g[e[0]][e[1]] = 2
}
for _, e := range walls {
g[e[0]][e[1]] = 2
}
dirs := [5]int{-1, 0, 1, 0, -1}
for _, e := range guards {
for k := 0; k < 4; k++ {
x, y := e[0], e[1]
a, b := dirs[k], dirs[k+1]
for x+a >= 0 && x+a < m && y+b >= 0 && y+b < n && g[x+a][y+b] < 2 {
x, y = x+a, y+b
g[x][y] = 1
}
}
}
for _, row := range g {
for _, v := range row {
if v == 0 {
ans++
}
}
}
return
}
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28 | function countUnguarded(m: number, n: number, guards: number[][], walls: number[][]): number {
const g: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
for (const [i, j] of guards) {
g[i][j] = 2;
}
for (const [i, j] of walls) {
g[i][j] = 2;
}
const dirs: number[] = [-1, 0, 1, 0, -1];
for (const [i, j] of guards) {
for (let k = 0; k < 4; ++k) {
let [x, y] = [i, j];
let [a, b] = [dirs[k], dirs[k + 1]];
while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
x += a;
y += b;
g[x][y] = 1;
}
}
}
let ans = 0;
for (const row of g) {
for (const v of row) {
ans += v === 0 ? 1 : 0;
}
}
return ans;
}
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28 | function countUnguarded(m, n, guards, walls) {
const g = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
for (const [i, j] of guards) {
g[i][j] = 2;
}
for (const [i, j] of walls) {
g[i][j] = 2;
}
const dirs = [-1, 0, 1, 0, -1];
for (const [i, j] of guards) {
for (let k = 0; k < 4; ++k) {
let [x, y] = [i, j];
let [a, b] = [dirs[k], dirs[k + 1]];
while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
x += a;
y += b;
g[x][y] = 1;
}
}
}
let ans = 0;
for (const row of g) {
for (const v of row) {
ans += v === 0 ? 1 : 0;
}
}
return ans;
}
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40 | impl Solution {
pub fn count_unguarded(m: i32, n: i32, guards: Vec<Vec<i32>>, walls: Vec<Vec<i32>>) -> i32 {
let m = m as usize;
let n = n as usize;
let mut g = vec![vec![0; n]; m];
for e in &guards {
g[e[0] as usize][e[1] as usize] = 2;
}
for e in &walls {
g[e[0] as usize][e[1] as usize] = 2;
}
let dirs = [-1, 0, 1, 0, -1];
for e in &guards {
let (x0, y0) = (e[0] as i32, e[1] as i32);
for k in 0..4 {
let (mut x, mut y) = (x0, y0);
let (a, b) = (dirs[k], dirs[k + 1]);
while x + a >= 0
&& x + a < m as i32
&& y + b >= 0
&& y + b < n as i32
&& g[(x + a) as usize][(y + b) as usize] < 2
{
x += a;
y += b;
g[x as usize][y as usize] = 1;
}
}
}
let mut ans = 0;
for row in g {
for v in row {
if v == 0 {
ans += 1;
}
}
}
ans
}
}
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33 | public class Solution {
public int CountUnguarded(int m, int n, int[][] guards, int[][] walls) {
int[,] g = new int[m, n];
foreach (var e in guards) {
g[e[0], e[1]] = 2;
}
foreach (var e in walls) {
g[e[0], e[1]] = 2;
}
int[] dirs = { -1, 0, 1, 0, -1 };
foreach (var e in guards) {
int x0 = e[0], y0 = e[1];
for (int k = 0; k < 4; ++k) {
int x = x0, y = y0;
int a = dirs[k], b = dirs[k + 1];
while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a, y + b] < 2) {
x += a;
y += b;
g[x, y] = 1;
}
}
}
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (g[i, j] == 0) {
++ans;
}
}
}
return ans;
}
}
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