Skip to content

463. Island Perimeter

Description

You are given row x col grid representing a map where grid[i][j] = 1 represents land and grid[i][j] = 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", meaning the water inside 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.

 

Example 1:

Input: grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]
Output: 16
Explanation: The perimeter is the 16 yellow stripes in the image above.

Example 2:

Input: grid = [[1]]
Output: 4

Example 3:

Input: grid = [[1,0]]
Output: 4

 

Constraints:

  • row == grid.length
  • col == grid[i].length
  • 1 <= row, col <= 100
  • grid[i][j] is 0 or 1.
  • There is exactly one island in grid.

Solutions

Solution 1

Thinking

The island is 4-connected; perimeter is \(4\) per land cell minus twice every shared edge. Checking all four neighbors needs bounds tests.

Add \(4\) for each land cell; if the cell below or to the right is also land, that edge is shared, so subtract \(2\). Looking only down and right counts each inner edge once.

No graph or DFS is required; one grid scan yields the perimeter.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
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;
}

Comments