Island Perimeter Leetcode

Posted 2020-09-22 璨璨要好好学习

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:

 

这道题我是按照遍历，看看如果周边有1的话周长就减1的做法做的。

public class Solution {
    public int islandPerimeter(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    count += 4 + isOne(grid, i, j);
                }
            }
        }
        return count;
    }
    private boolean inBound(int[][] grid, int x, int y) {
        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) {
            return false;
        }
        return true;
    }
    private int isOne(int[][] grid, int x, int y) {
        int count = 0;
        int[] dx = {0, 0, 1, -1};
        int[] dy = {1, -1, 0, 0};
        for (int i = 0; i < 4; i++) {
            int newX = x + dx[i];
            int newY = y + dy[i];
            if (inBound(grid, newX, newY) && grid[newX][newY] == 1) {
                count--;
            }
        }
        return count;
    }
}

做的可能有点复杂了，因为还有更好的做法。只要遍历右边和下边就可以了。最后结果是岛的个数 * 4 - 2 * 右边和下边的个数。因为右边和下边代表有两个岛要减一，所以乘2。

public class Solution {
    public int islandPerimeter(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    count += 4;
                    if (i + 1 < grid.length && grid[i + 1][j] == 1) {
                        count -= 2;
                    }
                    if (j + 1 < grid[0].length && grid[i][j + 1] == 1) {
                        count -= 2;
                    }
                }
            }
        }
        return count;
    }
}