200. Number of Islands-优快云博客

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.le 1:

Answer: 1

Answer: 3

When we met a ‘1’, the answer add 1, we also need to search all ‘1’ which connected to it directly or indirectly, and change it to ‘0’. And we can use DFS or BFS to search.

DFS
======
class Solution
{
public:
int numIslands(vector<vector<char>> &grid)
{
if(grid.size() == 0 || grid[0].size() == 0)
return 0;
```
     int res = 0;
     for(int i = 0; i < grid.size(); ++ i)
         for(int j = 0; j < grid[0].size(); ++ j)
             if(grid[i][j] == '1')
             {
                 ++ res;
                 DFS(grid, i, j);
             }
     return res;
 }
```
private:
void DFS(vector<vector<char>> &grid, int x, int y)
{
grid[x][y] = ‘0’;
if(x > 0 && grid[x - 1][y] == ‘1’)
DFS(grid, x - 1, y);
if(x < grid.size() - 1 && grid[x + 1][y] == ‘1’)
DFS(grid, x + 1, y);
if(y > 0 && grid[x][y - 1] == ‘1’)
DFS(grid, x, y - 1);
if(y < grid[0].size() - 1 && grid[x][y + 1] == ‘1’)
DFS(grid, x, y + 1);
}
};

BFS
======

class Solution
{
public:
int numIslands(vector<vector<char>> &grid)
{
if(grid.size() == 0 || grid[0].size() == 0)
return 0;

     int res = 0;
     for(int i = 0; i < grid.size(); ++ i)
         for(int j = 0; j < grid[0].size(); ++ j)
             if(grid[i][j] == '1')
             {
                 ++ res;
                 BFS(grid, i, j);
             }
     return res;
 }

private:
void BFS(vector<vector<char>> &grid, int x, int y)
{
queue<vector<int>> q;
q.push({x, y});
grid[x][y] = ‘0’;

     while(!q.empty())
     {
         x = q.front()[0], y = q.front()[1];
         q.pop();
         
         if(x > 0 && grid[x - 1][y] == '1')
         {
             q.push({x - 1, y});
             grid[x - 1][y] = '0';
         }
         if(x < grid.size() - 1 && grid[x + 1][y] == '1')
         {
             q.push({x + 1, y});
             grid[x + 1][y] = '0';
         }
         if(y > 0 && grid[x][y - 1] == '1')
         {
             q.push({x, y - 1});
             grid[x][y - 1] = '0';
         }
         if(y < grid[0].size() - 1 && grid[x][y + 1] == '1')
         {
             q.push({x, y + 1});
             grid[x][y + 1] = '0';
         }
     }
 }

};