<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 dx[4] = { 1,-1,0,0 };
    int dy[4] = { 0,0,1,-1 };
public:
    //创建dfs函数
    void dfs(vector<vector<int>>& grid, vector<vector<int>>& used, int& result, int row, int col) {
        //将当前岛屿标记为使用
        used[row][col] = 1;
        //初始化当前岛屿的周长
        int temp = 4;

        for (int i = 0; i < 4; i++) {
            int x = dx[i] + row;
            int y = dy[i] + col;    
            if (x >= 0 && y >= 0 && x < grid.size() && y < grid[0].size()) {
                //如果有相邻岛屿，则让当前岛屿周长减一
                if (grid[x][y] == 1) {
                    temp--;
                }
                //再dfs相邻岛屿
                if (grid[x][y] == 1 && used[x][y] == 0) {
                    dfs(grid, used, result, x, y);
                }
            }
        }
        //最后让结果加上当前岛屿的周长
        result += temp;
    }

    int islandPerimeter(vector<vector<int>>& grid) {
        //标记哪些岛屿被使用过
        vector<vector<int>> used(grid.size(), vector<int>(grid[0].size(), 0));
        //定义结果
        int result = 0;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == 1) {
                    //只用进入一块岛屿即可，因为所有的岛屿是连在一起的
                    dfs(grid, used, result, i, j);
                    return result;
                }
            }
        }

        return result;
    }
};
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