<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:
    const vector<int> dx = {-1, 1, 0, 0};
    const vector<int> dy = {0, 0, 1, -1};
    int n, m, vis[302][302], a[302][302];

    void dfs(int i, int j) {
        vis[i][j] = true;
        for (int z = 0; z < 4; z++) {
            int x = i + dx[z], y = j + dy[z];
            if (x < 0 || y < 0 || x >= n || y >= m || vis[x][y] || a[x][y] == 0) continue;
            dfs(x, y);
        }
    }

    int numIslands(vector<vector<char>>& grid) {
        n = grid.size(), m = grid[0].size();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = grid[i][j] - '0';
            }
        }

        int res = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!vis[i][j] && a[i][j] == 1) {
                    res++;
                    dfs(i, j);
                }
            }
        }
        return res;
    }
};

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