【leetcode】463. Island Perimeter

原题

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:

解析

求周长
给一个矩阵，0表示水，1表示陆地，求小岛的周长

思路

还是可以用BFS，或者直接遍历整个矩阵（缺点：矩阵很大陆地很小浪费效率）

BFS解法

    private int perimeter = 0;

    public int islandPerimeter(int[][] grid) {
        if (grid == null || grid.length <= 0 || grid[0] == null || grid[0].length <= 0) {
            return 0;
        }
        ArrayList<Point> isChecked = new ArrayList<>();
        for (int i = 0; i < grid.length * grid[0].length; i++) {
            //找到第一块陆地的位置
            if (grid[i / grid.length][i % grid.length] == 1) {
                getPerimeter(grid, isChecked, i / grid.length, i % grid.length);
                break;
            }
        }
        return perimeter;
    }

    /**
     * 判断边长递归方法
     * @param grid 矩阵
     * @param isChecked
     *@param i 位置row
     * @param j 位置col   @return 该位置是否陆地
     */
    private boolean getPerimeter(int[][] grid, ArrayList<Point> isChecked, int i, int j) {
        //若i,j是陆地，将它置为0，检测他的四周，是非陆地的边长+1，是陆地的递归getPerimeter
        if (i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1) {
            if (!isChecked.contains(new Point(i, j))) {
                isChecked.add(new Point(i, j));
                if (!getPerimeter(grid, isChecked, i - 1, j)) {
                    perimeter++;
                }
                if (!getPerimeter(grid, isChecked, i + 1, j)) {
                    perimeter++;
                }
                if (!getPerimeter(grid, isChecked, i, j - 1)) {
                    perimeter++;
                }
                if (!getPerimeter(grid, isChecked, i, j + 1)) {
                    perimeter++;
                }
            }
            return true;
        }
        return false;
    }

直接遍历解法

public int getPerimeterOther(int[][] grid) {
        int result = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    result += 4;
                    if (i > 0 && grid[i - 1][j] == 1) {
                        result -= 2;
                    }
                    if (j > 0 && grid[i][j - 1] == 1) {
                        result -= 2;
                    }
                }
            }
        }
        return result;
    }