LeetCode - Number of Islands

Number of Islands

2015.4.17 06:16

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

Solution1:

　　Solutiuon using DFS.

Accepted code:

 1 // 3CE, 1WA, 1AC, solution using DFS
 2 class Solution {
 3 public:
 4     Solution() {
 5         d[0][0] = -1;
 6         d[0][1] = 0;
 7         d[1][0] = +1;
 8         d[1][1] = 0;
 9         d[2][0] = 0;
10         d[2][1] = -1;
11         d[3][0] = 0;
12         d[3][1] = +1;
13     }
14 
15     int numIslands(vector<vector<char>> &grid) {
16         n = grid.size();
17         if (n == 0) {
18             return 0;
19         }
20         m = grid[0].size();
21         if (m == 0) {
22             return 0;
23         }
24         
25         int cc = 0;
26         int i, j;
27         
28         b.resize(n, vector<bool>(m, false));
29         for (i = 0; i < n; ++i) {
30             for (j = 0; j < m; ++j) {
31                 if (grid[i][j] == '1' && !b[i][j]) {
32                     DFS(i, j, grid);
33                     ++cc;
34                 }
35             }
36         }
37         b.clear();
38         
39         return cc;
40     }
41 protected:
42     int n, m;
43     vector<vector<bool> > b;
44     int d[4][2];
45     
46     void DFS(int x, int y, vector<vector<char> > &grid) {
47         b[x][y] = true;
48         int i;
49         int x1, y1;
50         
51         for (i = 0; i < 4; ++i) {
52             x1 = x + d[i][0];
53             y1 = y + d[i][1];
54             if (x1 < 0 || x1 > n - 1 || y1 < 0 || y1 > m - 1) {
55                 continue;
56             }
57             if (grid[x1][y1] == '0' || b[x1][y1]) {
58                 continue;
59             }
60             DFS(x1, y1, grid);
61         }
62     }
63 };

Solutiuon2:

　　Solution using BFS.

Accepted code:

 1 // 1AC, solution using BFS
 2 #include <queue>
 3 using namespace std;
 4 
 5 class Solution {
 6 public:
 7     Solution() {
 8         d[0][0] = -1;
 9         d[0][1] = 0;
10         d[1][0] = +1;
11         d[1][1] = 0;
12         d[2][0] = 0;
13         d[2][1] = -1;
14         d[3][0] = 0;
15         d[3][1] = +1;
16     }
17 
18     int numIslands(vector<vector<char>> &grid) {
19         n = grid.size();
20         if (n == 0) {
21             return 0;
22         }
23         m = grid[0].size();
24         if (m == 0) {
25             return 0;
26         }
27         
28         int cc = 0;
29         int i, j;
30         
31         b.resize(n, vector<bool>(m, false));
32         for (i = 0; i < n; ++i) {
33             for (j = 0; j < m; ++j) {
34                 if (grid[i][j] == '1' && !b[i][j]) {
35                     BFS(i, j, grid);
36                     ++cc;
37                 }
38             }
39         }
40         b.clear();
41         
42         return cc;
43     }
44 protected:
45     int n, m;
46     vector<vector<bool> > b;
47     int d[4][2];
48     
49     void BFS(int x, int y, vector<vector<char> > &grid) {
50         queue<int> q;
51         int i;
52         int x1, y1;
53         int p;
54         
55         q.push(x * m + y);
56         b[x][y] = true;
57         while (!q.empty()) {
58             p = q.front();
59             q.pop();
60             x = p / m;
61             y = p % m;
62             for (i = 0; i < 4; ++i) {
63                 x1 = x + d[i][0];
64                 y1 = y + d[i][1];
65                 if (x1 < 0 || x1 > n - 1 || y1 < 0 || y1 > m - 1) {
66                     continue;
67                 }
68                 if (grid[x1][y1] == '0' || b[x1][y1]) {
69                     continue;
70                 }
71                 q.push(x1 * m + y1);
72                 b[x1][y1] = true;
73             }
74         }
75     }
76 };