<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) {
        m = grid.size();
        n = grid.front().size();
        ans = 0;
        for (int ii = 0; ii < m; ++ii) {
            for (int jj = 0; jj < n; ++jj) {
                if (grid[ii][jj]==1) dfs(grid, ii, jj);
            }
        }
        return ans;
    }
    void dfs(vector<vector<int>>& grid, int x, int y) {
        grid[x][y] = -1;
        int diff_x[4] = {0,1,0,-1};
        int diff_y[4] = {1,0,-1,0};
        for (int ii = 0; ii < 4; ++ii) {
            int xx = x + diff_x[ii];
            int yy = y + diff_y[ii];
            if (xx>=0 && xx <m && yy>=0 && yy<n) {
                if (grid[xx][yy]==1) dfs(grid,xx,yy);
                else if (grid[xx][yy]==0) ans++;
            }
            else ans++;
        }
        return;
    }
private:
    int m, n;
    int ans;
};
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