200. Number of Islands

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

题意：统计矩阵中岛屿的个数。

思路：dfs。

class Solution {public:int numIslands(vector<vector<char>>& grid) {const int m = grid.size();if (m == 0)return 0;const int n = grid[0].size();int res = 0;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (grid[i][j] == '1'){res++;dfs(grid, i, j, res+1);}}}return res;}private:void dfs(vector<vector<char>>& grid, int i, int j, int num){if (i >= 0 && i < grid.size() && j >= 0 && j < grid[0].size()){if (grid[i][j] == '1'){grid[i][j] = ('0' + num);dfs(grid, i - 1, j, num);    dfs(grid, i + 1, j, num);    dfs(grid, i, j - 1, num);    dfs(grid, i, j + 1, num);}}}};

二刷：

用标志数组的思路实现。因为如果岛屿的数量大于255，上述算法不太适应。但是，上述算法不需要开辟额外空间，值得借鉴。15.62%

class Solution {public:int numIslands(vector<vector<char>>& grid) {if (grid.empty())return 0;const int LINE = grid.size();const int COLUMN = grid[0].size();vector<vector<bool>> visited(LINE, vector<bool>(COLUMN, false));int res = 0;;for (int i = 0; i < LINE; i++){for (int j = 0; j < COLUMN; j++){if (grid[i][j] == '1'  &&  !visited[i][j]){dfs(grid, visited, i, j);res++;}}}return res;}private:void dfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int i, int j){if (i<0 || i>=grid.size())return;if (j < 0 || j >= grid[0].size())return;if (grid[i][j] == '1'  &&  !visited[i][j]){visited[i][j] = true;dfs(grid, visited, i + 1, j);dfs(grid, visited, i, j + 1);dfs(grid, visited, i - 1, j);dfs(grid, visited, i, j - 1);}}};