leetcode_Number of Islands_medium--dfs

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

题目：其实本质上类似于，给了一个矩形点阵，找连通域个数。

方法：深度优先搜索(dfs)

class Solution {
private:
int res=0;
int dfs(vector<vector<char>> &grid,vector<vector<char>> &vis,int i,int j)
{
vis[i][j]=1;//已访问
if(i>0)
if(0 == vis[i-1][j] && '1' == grid[i-1][j])
dfs(grid,vis,i-1,j);//访问正上方节点
if(j < grid[0].size()-1)
if(0 == vis[i][j+1] && '1' == grid[i][j+1])
dfs(grid,vis,i,j+1);//访问正右边节点
if(i < grid.size()-1)
if(0 == vis[i+1][j] && '1' == grid[i+1][j])
dfs(grid,vis,i+1,j);//访问正下方节点
if(j>0)
if(0 == vis[i][j-1] && '1'== grid[i][j-1])
dfs(grid,vis,i,j-1);//访问正左边节点
return 0;
}
public:
int numIslands(vector<vector<char>>& grid) {
if(grid.empty() == true)
return 0;
int rows=grid.size(),cols=grid[0].size();
vector<char> tmp(cols,0);
vector<vector<char> > vis(rows,tmp);
for(int i=0; i<rows; i++)//遍历每行
for(int j=0; j<cols; j++)//遍历每列
{
if(0 == vis[i][j] && '1' == grid[i][j])//没有被访问过，且节点为1才进行搜索
{
res++;
dfs(grid,vis,i,j);
}
}
return res;
}
};