# [463. 岛屿的周长](https://leetcode.cn/problems/island-perimeter)

[English Version](/solution/0400-0499/0463.Island%20Perimeter/README_EN.md)

## 题目描述

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<p>给定一个 <code>row x col</code> 的二维网格地图 <code>grid</code> ，其中：<code>grid[i][j] = 1</code> 表示陆地， <code>grid[i][j] = 0</code> 表示水域。</p>

<p>网格中的格子 <strong>水平和垂直</strong> 方向相连（对角线方向不相连）。整个网格被水完全包围，但其中恰好有一个岛屿（或者说，一个或多个表示陆地的格子相连组成的岛屿）。</p>

<p>岛屿中没有“湖”（“湖” 指水域在岛屿内部且不和岛屿周围的水相连）。格子是边长为 1 的正方形。网格为长方形，且宽度和高度均不超过 100 。计算这个岛屿的周长。</p>

<p> </p>

<p><strong>示例 1：</strong></p>

<p><img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0463.Island%20Perimeter/images/island.png" /></p>

<pre>
<strong>输入：</strong>grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]
<strong>输出：</strong>16
<strong>解释：</strong>它的周长是上面图片中的 16 个黄色的边</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>grid = [[1]]
<strong>输出：</strong>4
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>grid = [[1,0]]
<strong>输出：</strong>4
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>row == grid.length</code></li>
	<li><code>col == grid[i].length</code></li>
	<li><code>1 <= row, col <= 100</code></li>
	<li><code>grid[i][j]</code> 为 <code>0</code> 或 <code>1</code></li>
</ul>

## 解法

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遍历二维数组

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### **Python3**

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```python
class Solution:
    def islandPerimeter(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        ans = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    ans += 4
                    if i < m - 1 and grid[i + 1][j] == 1:
                        ans -= 2
                    if j < n - 1 and grid[i][j + 1] == 1:
                        ans -= 2
        return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
    public int islandPerimeter(int[][] grid) {
        int ans = 0;
        int m = grid.length;
        int n = grid[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    ans += 4;
                    if (i < m - 1 && grid[i + 1][j] == 1) {
                        ans -= 2;
                    }
                    if (j < n - 1 && grid[i][j + 1] == 1) {
                        ans -= 2;
                    }
                }
            }
        }
        return ans;
    }
}
```

### **TypeScript**

```ts
function islandPerimeter(grid: number[][]): number {
    let m = grid.length,
        n = grid[0].length;
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            let top = 0,
                left = 0;
            if (i > 0) {
                top = grid[i - 1][j];
            }
            if (j > 0) {
                left = grid[i][j - 1];
            }
            let cur = grid[i][j];
            if (cur != top) ++ans;
            if (cur != left) ++ans;
        }
    }
    // 最后一行， 最后一列
    for (let i = 0; i < m; ++i) {
        if (grid[i][n - 1] == 1) ++ans;
    }
    for (let j = 0; j < n; ++j) {
        if (grid[m - 1][j] == 1) ++ans;
    }
    return ans;
}
```

### **C++**

```cpp
class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ans += 4;
                    if (i < m - 1 && grid[i + 1][j] == 1) ans -= 2;
                    if (j < n - 1 && grid[i][j + 1] == 1) ans -= 2;
                }
            }
        }
        return ans;
    }
};
```

### **Go**

```go
func islandPerimeter(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	ans := 0
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				ans += 4
				if i < m-1 && grid[i+1][j] == 1 {
					ans -= 2
				}
				if j < n-1 && grid[i][j+1] == 1 {
					ans -= 2
				}
			}
		}
	}
	return ans
}
```

### **...**

```

```

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