【LeetCode】200. Number of Islands

题目：

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

分析：

这道题的要求是求一个矩阵中有多少块被‘0’包围的‘1’区域。我的想法是使用DFS算法。设一个变量sum来记录”小岛“的数目，通过双重循环遍历整个矩阵，若遍历到‘1’则通过深度优先遍历将与其为一个区域的‘1’标记。这样调用DFS的次数就是”小岛“的数目。

代码：

class Solution {
public:
void DFS(vector<vector<char>>& grid,int i,int j)
{
grid[i][j] = 'A';    //将遍历到的元素进行标记，以免重复遍历
if(i>0)
{
if(grid[i-1][j] == '1') //上
DFS(grid,i-1,j);
}
if(i<grid.size()-1)//下
{
if(grid[i+1][j] == '1')
DFS(grid,i+1,j);
}
if(j>0)
{
if(grid[i][j-1] == '1')//左
DFS(grid,i,j-1);
}
if(j<grid[0].size()-1)//右
{
if(grid[i][j+1] == '1')
DFS(grid,i,j+1);
}
}
int numIslands(vector<vector<char>>& grid) {
int sum =0;
int n = grid.size();
if(n == 0)
return 0;
int m = grid[0].size();
for(int i = 0;i<n;i++)
{
for(int j = 0;j<m;j++)
{
if(grid[i][j] == '1')
{
DFS(grid,i,j);
sum++;
}

}
}
return sum;
}
};