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 };
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