题目描述
在由 1 x 1 方格组成的 n x n 网格 grid 中，每个 1 x 1 方块由 '/'、'\' 或空格构成。这些字符会将方块划分为一些共边的区域。
给定网格 grid 表示为一个字符串数组，返回 区域的数量 。
请注意，反斜杠字符是转义的，因此 '\' 用 '\\' 表示。
示例 1：
输入：grid = [" /","/ "]
输出：2
示例 2：
输入：grid = [" /"," "]
输出：1
示例 3：
输入：grid = ["/\\","\\/"]
输出：5
解释：回想一下，因为 \ 字符是转义的，所以 "/\\" 表示 /\，而 "\\/" 表示 \/。
提示：
n == grid.length == grid[i].length1 <= n <= 30grid[i][j] 是 '/'、'\'、或 ' '
解法
方法一
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35 | class Solution:
def regionsBySlashes(self, grid: List[str]) -> int:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
def union(a, b):
pa, pb = find(a), find(b)
if pa != pb:
p[pa] = pb
nonlocal size
size -= 1
n = len(grid)
size = n * n * 4
p = list(range(size))
for i, row in enumerate(grid):
for j, v in enumerate(row):
k = i * n + j
if i < n - 1:
union(4 * k + 2, (k + n) * 4)
if j < n - 1:
union(4 * k + 1, (k + 1) * 4 + 3)
if v == '/':
union(4 * k, 4 * k + 3)
union(4 * k + 1, 4 * k + 2)
elif v == '\\':
union(4 * k, 4 * k + 1)
union(4 * k + 2, 4 * k + 3)
else:
union(4 * k, 4 * k + 1)
union(4 * k + 1, 4 * k + 2)
union(4 * k + 2, 4 * k + 3)
return size
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54 | class Solution {
private int[] p;
private int size;
public int regionsBySlashes(String[] grid) {
int n = grid.length;
size = n * n * 4;
p = new int[size];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int k = i * n + j;
if (i < n - 1) {
union(4 * k + 2, (k + n) * 4);
}
if (j < n - 1) {
union(4 * k + 1, (k + 1) * 4 + 3);
}
char v = grid[i].charAt(j);
if (v == '/') {
union(4 * k, 4 * k + 3);
union(4 * k + 1, 4 * k + 2);
} else if (v == '\\') {
union(4 * k, 4 * k + 1);
union(4 * k + 2, 4 * k + 3);
} else {
union(4 * k, 4 * k + 1);
union(4 * k + 1, 4 * k + 2);
union(4 * k + 2, 4 * k + 3);
}
}
}
return size;
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
private void union(int a, int b) {
int pa = find(a);
int pb = find(b);
if (pa == pb) {
return;
}
p[pa] = pb;
--size;
}
}
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45 | class Solution {
public:
vector<int> p;
int size;
int regionsBySlashes(vector<string>& grid) {
int n = grid.size();
size = n * n * 4;
p.resize(size);
for (int i = 0; i < size; ++i) p[i] = i;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int k = i * n + j;
if (i < n - 1) merge(4 * k + 2, (k + n) * 4);
if (j < n - 1) merge(4 * k + 1, (k + 1) * 4 + 3);
char v = grid[i][j];
if (v == '/') {
merge(4 * k, 4 * k + 3);
merge(4 * k + 1, 4 * k + 2);
} else if (v == '\\') {
merge(4 * k, 4 * k + 1);
merge(4 * k + 2, 4 * k + 3);
} else {
merge(4 * k, 4 * k + 1);
merge(4 * k + 1, 4 * k + 2);
merge(4 * k + 2, 4 * k + 3);
}
}
}
return size;
}
void merge(int a, int b) {
int pa = find(a);
int pb = find(b);
if (pa == pb) return;
p[pa] = pb;
--size;
}
int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};
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46 | func regionsBySlashes(grid []string) int {
n := len(grid)
size := n * n * 4
p := make([]int, size)
for i := range p {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
union := func(a, b int) {
pa, pb := find(a), find(b)
if pa == pb {
return
}
p[pa] = pb
size--
}
for i, row := range grid {
for j, v := range row {
k := i*n + j
if i < n-1 {
union(4*k+2, (k+n)*4)
}
if j < n-1 {
union(4*k+1, (k+1)*4+3)
}
if v == '/' {
union(4*k, 4*k+3)
union(4*k+1, 4*k+2)
} else if v == '\\' {
union(4*k, 4*k+1)
union(4*k+2, 4*k+3)
} else {
union(4*k, 4*k+1)
union(4*k+1, 4*k+2)
union(4*k+2, 4*k+3)
}
}
}
return size
}
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