695 Max Area of Island

695 Max Area of Island

1. 题目：

Given a non-empty 2D arraygridof 0's and 1's, anislandis a group of1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[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]]

Given the above grid, return

6

Note the answer is not 11, because the island must be connected 4-directionally.

2. 思路

(1) 利用Union Find的模板, 每个set都有对应的size

(2) DFS

(3) BFS

3. Implementation

1. Union Find解法

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int m = grid.length, n = grid[0].length;
        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        UnionFind uf = new UnionFind(grid);

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    for (int[] direction : directions) {
                        int nextRow = i + direction[0];
                        int nextCol = j + direction[1];

                        if (isValid(nextRow, nextCol, grid)) {
                            int id1 = i * n + j;
                            int id2 = nextRow * n + nextCol;
                            uf.union(id1, id2);
                        }
                    }
                }
            }
        }
        return uf.maxArea();
    }

    public boolean isValid(int row, int col, int[][] grid) {
        return row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && grid[row][col] == 1;
    }

    class UnionFind {
        int[] sets;
        int[] size;
        int m, n, count;

        public UnionFind(int[][] grid) {
            m = grid.length;
            n = grid[0].length;
            sets = new int[m * n];
            size = new int[m * n];

            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (grid[i][j] == 1) {
                        int id = i * n + j;
                        sets[id] = id;
                        size[id] = 1;
                        ++count;
                    }
                }
            }
        }

        public int find(int node) {
            while (node != sets[node]) {
                node = sets[node];
            }
            return node;
        }

        public void union(int i, int j) {
            int node1 = find(i);
            int node2 = find(j);

            if (node1 == node2) {
                return;
            }

            if (size[node1] < size[node2]) {
                sets[node1] = node2;
                size[node2] += size[node1];
            }
            else {
                sets[node2] = node1;
                size[node1] += size[node2];
            }

            --count;
        }

        public int maxArea() {
            int res = 0;
            for (int i = 0; i < size.length; i++) {
                res = Math.max(res, size[i]);
            }
            return res;
        }
    }
}