<p>You are given <code>row x col</code> <code>grid</code> representing a map where <code>grid[i][j] = 1</code> represents&nbsp;land and <code>grid[i][j] = 0</code> represents water.</p>

<p>Grid cells are connected <strong>horizontally/vertically</strong> (not diagonally). The <code>grid</code> is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells).</p>

<p>The island doesn&#39;t have &quot;lakes&quot;, meaning the water inside isn&#39;t connected to the water around the island. One cell is a square with side length 1. The grid is rectangular, width and height don&#39;t exceed 100. Determine the perimeter of the island.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2018/10/12/island.png" style="width: 221px; height: 213px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The perimeter is the 16 yellow stripes in the image above.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0]]
<strong>Output:</strong> 4
</pre>

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

<ul>
	<li><code>row == grid.length</code></li>
	<li><code>col == grid[i].length</code></li>
	<li><code>1 &lt;= row, col &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is <code>0</code> or <code>1</code>.</li>
	<li>There is exactly one island in <code>grid</code>.</li>
</ul>

class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        // dfs
        // hit water  or a border, add 1
        int perim = 0;
        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size()));
        for(int i = 0; i < grid.size(); i++)
        {
            for(int j = 0; j < grid[0].size(); j++)
            {
                if(grid[i][j] == 1)
                    dfs(grid, i, j, visited, perim);
            }
        }
        return perim;
    }

    void dfs(vector<vector<int>>& grid, int i, int j, vector<vector<bool>>& visited, int& perim)
    {
        if(i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size() || grid[i][j] == 0)
        {
            perim++;
            return;
        }
        if(visited[i][j] == true)
            return;
        visited[i][j] = true;
        dfs(grid, i + 1, j, visited, perim);
        dfs(grid, i - 1, j, visited, perim);
        dfs(grid, i, j + 1, visited, perim);
        dfs(grid, i, j - 1, visited, perim);
    }
};
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