<p>Given an <code>m x n</code> 2D binary grid <code>grid</code> which represents a map of <code>&#39;1&#39;</code>s (land) and <code>&#39;0&#39;</code>s (water), return <em>the number of islands</em>.</p>

<p>An <strong>island</strong> 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.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;]
]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;]
]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>grid[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    int x[4] = {-1,1,0,0};
    int y[4] = {0,0,-1,1};
    void dfs(vector<vector<char>>& grid,vector<vector<bool>>& visited,int i,int j,int m,int n){
        visited[i][j] = 1;
        for(int k=0;k<4;k++){
            if(i+x[k] >=0 && i+x[k] < m && j+y[k] >=0 && j+y[k] < n && !visited[i+x[k]][j+y[k]] && grid[i+x[k]][j+y[k]] == '1'){
                dfs(grid,visited,i+x[k],j+y[k],m,n);
            }
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();

        vector<vector<bool>> visited(m,vector<bool>(n,0));
        int ans = 0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(!visited[i][j] && grid[i][j] == '1'){
                    dfs(grid,visited,i,j,m,n);
                    ans++;
                }
            }
        }
        return ans;
    }
};
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