1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
|
// 1254. Number of Closed Islands
#include "leetcode.h"
/*
* given 2d 'grid' consists of 0's (land) and 1's (water). an island is a
* maximal four directionally connected group of 0's and a closed island is an
* island totally (all left, top, right, bottom) surrouned by 1's. return the
* number of closed islands.
*/
class Solution {
int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};
bool is_valid(vvd(int) & grid, int i, int j, int r, int c) {
if (i >= 0 && i < r && j >= 0 && j < c && grid[i][j] == 0)
return true;
else
return false;
}
void dfs(vvd(int) & grid, int i, int j, int r, int c) {
grid[i][j] = 1;
for (int x = 0; x < 4; ++x) {
int new_x = i + dx[x];
int new_y = i + dy[x];
if (is_valid(grid, new_x, new_y, r, c))
dfs(grid, new_x, new_y, r, c);
}
}
public:
int closedIsland(vvd(int) & grid) {
int r = grid.size(), c = grid[0].size(), ans = 0;
for (int i = 0; i < r; ++i)
for (int j = 0; j < c; ++j)
if (i * j == 0 || i == r - 1 || j == c - 1)
if (grid[i][j] == 0)
dfs(grid, i, j, r, c);
for (int i = 0; i < r; ++i)
for (int j = 0; j < c; ++j)
if (grid[i][j] == 0) {
ans++;
dfs(grid, i, j, r, c);
}
return ans;
}
};
int main() {
Solution obj;
vvd(int) g1 = {{1, 1, 1, 1, 1, 1, 1, 0},
{1, 0, 0, 0, 0, 1, 1, 0},
{1, 0, 0, 0, 0, 1, 0, 1},
{1, 1, 1, 1, 1, 1, 1, 0}};
vvd(int) g2 = {{0, 0, 1, 0, 0}, {0, 1, 0, 1, 0}};
vvd(int)
g3 = {{1, 1, 1, 1, 1, 1, 1}, {1, 0, 0, 0, 0, 0, 1}, {1, 0, 1, 1, 1, 0, 1},
{1, 0, 1, 0, 1, 0, 1}, {1, 0, 1, 1, 1, 0, 1}, {1, 0, 0, 0, 0, 0, 1},
{1, 1, 1, 1, 1, 1, 1}};
printf("%d\n", obj.closedIsland(g1)); // expect: 2
printf("%d\n", obj.closedIsland(g2)); // expect: 1
printf("%d\n", obj.closedIsland(g3)); // expect: 2
}
|