LintCode 1225: Island Perimeter-优快云博客

本文链接：https://blog.youkuaiyun.com/roufoo/article/details/106550521

本文介绍了一种计算二维网格中岛屿周长的方法。该方法通过遍历每个单元格，检查其周围环境来确定岛屿的边界，从而计算出岛屿的总周长。

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1225. 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:

Input test data (one parameter per line)How to understand a testcase?

解法1：

class Solution {
public:
    /**
     * @param grid: a 2D array
     * @return: the perimeter of the island
     */
    int islandPerimeter(vector<vector<int>> &grid) {
        int nRow = grid.size();
        if (nRow == 0) return 0;
        int nCol = grid[0].size();
        if (nCol == 0) return 0;
        
        int perimeter = 0;
        for (int i = 0; i < nRow; ++i) {
            for (int j = 0; j < nCol; ++j) {
                if (grid[i][j] == 1) {
                    if (i == 0) perimeter++;
                    if (i == nRow - 1) perimeter++;
                    if (j == 0) perimeter++;
                    if (j == nCol - 1) perimeter++;
                    if (i > 0 && grid[i - 1][j] == 0) perimeter++;
                    if (i < nRow - 1 && grid[i + 1][j] == 0) perimeter++;
                    if (j > 0 && grid[i][j - 1] == 0) perimeter++;
                    if (j < nCol - 1 && grid[i][j + 1] == 0) perimeter++;
                }
            }
        }
        
        return perimeter;
    }
};