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This is a typical problem about searching. In fact, you can use either BFS or DFS for it. Personally, I use BFS because I think it is more intuitive and easy to implement.
The idea is fairly simple. We maintain another 2-dimensional vector visited to mark whether the corresponding position in grid has been visited and a variable count for the number of islands (initialized to be 0). Then we visit grid from left to right and top to bottom. Each time when we see a non-visited 1 in grid, we add the number of islands by 1 and mark all the connected 4-neighbors (BFS is applied here to get neighbors, neighbors of neighbors, neighbors of neighbors of neighbors, ...) of the non-visited 1 (including itself) as visited.
Note that a lot of the solutions in the discussion forum directly changes 1 in grid to 0 instead of using another 2-dimensional vector visited. However, I personally prefer to keep the input unchanged in a function with a return value.
The code is as follows.
1 class Solution { 2 public: 3 int numIslands(vector<vector<char>>& grid) { 4 if (grid.empty()) return 0; 5 int m = grid.size(), n = grid[0].size(); 6 vector<vector<bool> > visited(m, vector<bool> (n, false)); 7 int count = 0; 8 for (int i = 0; i < m; i++) { 9 for (int j = 0; j < n; j++) { 10 if (grid[i][j] == ‘1‘ && !visited[i][j]){ 11 count++; 12 markIslands(grid, visited, i, j); 13 } 14 } 15 } 16 return count; 17 } 18 private: 19 void markIslands(vector<vector<char> >& grid, vector<vector<bool> >& visited, int row, int col) { 20 visited[row][col] = true; 21 if (row - 1 >= 0 && grid[row - 1][col] == ‘1‘ && !visited[row - 1][col]) 22 markIslands(grid, visited, row - 1, col); 23 if (row + 1 < grid.size() && grid[row + 1][col] == ‘1‘ && !visited[row + 1][col]) 24 markIslands(grid, visited, row + 1, col); 25 if (col - 1 >= 0 && grid[row][col - 1] == ‘1‘ && !visited[row][col - 1]) 26 markIslands(grid, visited, row, col - 1); 27 if (col + 1 < grid[0].size() && grid[row][col + 1] == ‘1‘ && !visited[row][col + 1]) 28 markIslands(grid, visited, row, col + 1); 29 } 30 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4596231.html
