[LeetCode] Island Perimeter 島嶼周長

 

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.

Example:

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

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

 

這道題給了我們一個格子圖，若干連在一起的格子形成了一個小島，規定了圖中只有一個相連的島，且島中沒有湖，讓我們求島的周長。我們知道一個格子有四條邊，但是當兩個格子相鄰，周圍為6，若某個格子四周都有格子，那么這個格子一條邊都不算在周長里。那么我們怎么統計出島的周長呢？第一種方法，我們對於每個格子的四條邊分別來處理，首先看左邊的邊，只有當左邊的邊處於第一個位置或者當前格子的左面沒有島格子的時候，左邊的邊計入周長。其他三條邊的分析情況都跟左邊的邊相似，這里就不多敘述了，參見代碼如下：

 

解法一：

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;
    }
};

 

下面這種方法對於每個島嶼格子先默認加上四條邊，然后檢查其左面和上面是否有島嶼格子，有的話分別減去兩條邊，這樣也能得到正確的結果，參見代碼如下：

 

解法二：

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;
    }
};

 

參考資料：

https://discuss.leetcode.com/topic/68983/java-9-line-solution-add-4-for-each-land-and-remove-2-for-each-internal-edge

 

LeetCode All in One 題目講解匯總(持續更新中...)