zoukankan      html  css  js  c++  java
  • [leetcode-463-Island Perimeter]

    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:
    

    参考的大牛的思路:

    1. find how many 1 in the map. If without the consideration of surrounding cells, the total perimeter should be the total amount of 1 times 4.
    2. find how many cell walls that connect with both lands. We need to deduct twice of those lines from total perimeter
    int islandPerimeter(vector<vector<int>>& grid)
         {
             int count = 0, repeat = 0;
             for (int i = 0; i < grid.size(); i++)
             {
                 for (int j = 0; j < grid[i].size(); j++)
                 {
                     if (grid[i][j] == 1)
                     {
                         count++;
                         if (i != 0 && grid[i - 1][j] == 1) repeat++;
                         if (j != 0 && grid[i][j - 1] == 1) repeat++;
                     }
                 }
             }
             return 4 * count - repeat * 2;
         }

    参考:

    https://discuss.leetcode.com/topic/68845/c-solution-with-explanation

  • 相关阅读:
    Spoj 2798 Qtree3
    [HAOI2015]树上操作
    Grass Planting
    [ZJOI2008] 树的统计Count
    Spoj375 Qtree--树链剖分
    [HNOI2012]永无乡
    雨天的尾巴
    temp
    线段树动态开点之逆序对
    线段树动态开点
  • 原文地址:https://www.cnblogs.com/hellowooorld/p/6842902.html
Copyright © 2011-2022 走看看