827. 最大人工岛

给你一个大小为 n x n 二进制矩阵 grid 。最多 只能将一格 0 变成 1 。

返回执行此操作后，grid 中最大的岛屿面积是多少？

岛屿 由一组上、下、左、右四个方向相连的 1 形成。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/making-a-large-island
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {

    private int[][] directions = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

    private int dfs(int[][] grid, int x, int y, int number) {
        int ans = 1;
        grid[x][y] = number;
        for (int i = 0; i < directions.length; ++i) {
            int nx = x + directions[i][0];
            int ny = y + directions[i][1];
            if (nx >= 0 && nx < grid.length && ny >= 0 && ny < grid[0].length && grid[nx][ny] == 1) {
                ans += dfs(grid, nx, ny, number);
            }
        }
        return ans;
    }

    public int largestIsland(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int ans = 0;
        int number = 2;
        Map<Integer, Integer> numMap = new HashMap<>();
        for (int i = 0; i < grid.length; ++i) {
            for (int j = 0; j < grid[0].length; ++j) {
                if (grid[i][j] == 1) {
                    int num = dfs(grid, i, j, number);
                    ans = Math.max(ans, num);
                    numMap.put(number++, num);
                }
            }
        }
        for (int i = 0; i < grid.length; ++i) {
            for (int j = 0; j < grid[0].length; ++j) {
                if (grid[i][j] == 0) {
                    Set<Integer> set = new HashSet<Integer>() {{
                        add(0);
                    }};
                    int num = 1;
                    for (int k = 0; k < directions.length; ++k) {
                        int x = i + directions[k][0];
                        int y = j + directions[k][1];
                        if (x >= 0 && x < grid.length && y >= 0 && y < grid[0].length && !set.contains(grid[x][y])) {
                            set.add(grid[x][y]);
                            num += numMap.get(grid[x][y]);
                        }
                    }
                    ans = Math.max(ans, num);
                }
            }
        }
        return ans;
    }
}