# [LeetCode] 463. Island Perimeter 岛的周长

2021年09月15日 阅读数：1

You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn't have "lakes" (water inside that isn'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't exceed 100. Determine the perimeter of the island.html

Example:java

```[[0,1,0,0],
[1,1,1,0],
[0,1,0,0],
[1,1,0,0]]

Explanation: The perimeter is the 16 yellow stripes in the image below:
```

Java:url

```public class Solution {
public int islandPerimeter(int[][] grid) {
int islands = 0, neighbours = 0;

for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == 1) {
islands++; // count islands
if (i < grid.length - 1 && grid[i + 1][j] == 1) neighbours++; // count down neighbours
if (j < grid[i].length - 1 && grid[i][j + 1] == 1) neighbours++; // count right neighbours
}
}
}

return islands * 4 - neighbours * 2;
}
}
```

Python:orm

```def islandPerimeter(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
ans = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
ans += 4
if i > 0 and grid[i-1][j] == 1:
ans -= 2
if j > 0 and grid[i][j-1] == 1:
ans -= 2
return ans　　```

Python:htm

```class Solution(object):
def islandPerimeter(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
if not grid:
return 0

adjacent = (i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1),
res = 0
if x < 0 or y < 0 or x == len(grid) or y == len(grid[0]) or grid[x][y] == 0:
res += 1
return res

count = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
return count　　```

Python:blog

```class Solution:
def islandPerimeter(self, grid):
if not grid:
return 0

N = len(grid)
M = len(grid[0])
ans = 0
for i in range(N):
for j in range(M):
for k in [(-1, 0), (0, 1), (1, 0), (0, -1)]:
if i+k[0] < 0 or i + k[0] >= N or j+k[1] < 0 or j+k[1] >= M or grid[i+k[0]][j+k[1]] == 0:
ans += 1
return ans　　```

C++:ip

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

C++:

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

Followup: 若是不止一个岛屿