_695_maxAreaOfIsland.java

package pp.arithmetic.leetcode;

/**
 * Created by wangpeng on 2019-02-21.
 * <p>
 * 695. 岛屿的最大面积
 * <p>
 * 给定一个包含了一些 0 和 1的非空二维数组 grid , 一个 岛屿 是由四个方向 (水平或垂直) 的 1 (代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。
 * <p>
 * 找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为0。)
 * <p>
 * 示例 1:
 * <p>
 * [
 * [0,0,1,0,0,0,0,1,0,0,0,0,0],
 * [0,0,0,0,0,0,0,1,1,1,0,0,0],
 * [0,1,1,0,1,0,0,0,0,0,0,0,0],
 * [0,1,0,0,1,1,0,0,1,0,1,0,0],
 * [0,1,0,0,1,1,0,0,1,1,1,0,0],
 * [0,0,0,0,0,0,0,0,0,0,1,0,0],
 * [0,0,0,0,0,0,0,1,1,1,0,0,0],
 * [0,0,0,0,0,0,0,1,1,0,0,0,0]
 * ]
 * 对于上面这个给定矩阵应返回 6。注意答案不应该是11，因为岛屿只能包含水平或垂直的四个方向的‘1’。
 * <p>
 * 示例 2:
 * <p>
 * [[0,0,0,0,0,0,0,0]]
 * 对于上面这个给定的矩阵, 返回 0。
 * <p>
 * 注意: 给定的矩阵grid 的长度和宽度都不超过 50。
 *
 * @see <a href="https://leetcode-cn.com/problems/max-area-of-island/">max-area-of-island</a>
 */
public class _695_maxAreaOfIsland {
    public static void main(String[] args) {
        int[][] grid = new int[][]{
                {0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
                {0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0},
                {0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},
                {0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0},
                {0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0},
                {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
                {0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0},
                {0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0}
        };
        System.out.println(maxAreaOfIsland(grid));
        System.out.println(maxAreaOfIsland(new int[][]{
                {1, 1}, {1, 0}
        }));
        System.out.println(maxAreaOfIsland(new int[][]{
                {1, 1, 1}, {1, 0, 0}
        }));
    }

    /**
     * 时间复杂度：O(M*N)
     * <p>
     * 执行用时: 41 ms, 在Max Area of Island的Java提交中击败了64.67% 的用户
     * 内存消耗: 37.9 MB, 在Max Area of Island的Java提交中击败了56.28% 的用户
     *
     * @param grid
     * @return
     */
    public static int maxAreaOfIsland(int[][] grid) {
        if (grid.length == 0) return 0;
        int max = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                max = Math.max(max, area(grid, i, j));
            }
        }
        return max;
    }

    private static int area(int[][] grid, int i, int j) {
        if (i >= 0 && i < grid.length && j >= 0 && j < grid[i].length && grid[i][j] != 0) {
            grid[i][j] = 0;
            return 1 + area(grid, i + 1, j) + area(grid, i - 1, j) + area(grid, i, j + 1) + area(grid, i, j - 1);
        }
        return 0;
    }
}