• LeetCode 463 Island Perimeter


    Problem:

    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:
    

    Summary:

    求正方形拼接在一起后的周长。

    Solution:

    1. 通过枚举可以得出规律,a个边长为1的独立正方形周长为4a,每当两个正方形拼接在一起产生一条重复边,周长减2。

    故有若干个正方形拼接在一起产生n条重复边时,周长为4a-2n。

    代码如下:

     1 class Solution {
     2 public:
     3     int islandPerimeter(vector<vector<int>>& grid) {
     4         int row = grid.size(), col = grid[0].size();
     5         int cnt = 0, repeat = 0;
     6         
     7         for (int i = 0; i < row; i++) {
     8             for (int j = 0; j < col; j++) {
     9                 if (grid[i][j]) {
    10                     cnt++;
    11                     if (i && grid[i - 1][j]) repeat++;
    12                     if (j && grid[i][j - 1]) repeat++;
    13                 }
    14             }
    15         }
    16         
    17         return cnt * 4 - repeat * 2;
    18     }
    19 };

    2. 分别计算每个方格的四个边

     1 class Solution {
     2 public:
     3     int islandPerimeter(vector<vector<int>>& grid) {
     4         int row = grid.size(), col = grid[0].size();
     5         int sum = 0;
     6         
     7         for (int i = 0; i < row; i++) {
     8             for (int j = 0; j < col; j++) {
     9                 if (grid[i][j]) {
    10                     if (!i || !grid[i - 1][j]) sum++;
    11                     if (i == row - 1 || !grid[i + 1][j]) sum++;
    12                     if (!j || !grid[i][j - 1]) sum++;
    13                     if (j == col - 1 || !grid[i][j + 1]) sum++;
    14                 }
    15             }
    16         }
    17         
    18         return sum;
    19     }
    20 };
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  • 原文地址:https://www.cnblogs.com/VickyWang/p/6186425.html
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