695 Max Area of Island

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695. Max Area of Island​
1. Question
Given a non-empty 2D arraygridof 0's and 1's, an island is 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:
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[[0,0,1,0,0,0,0,1,0,0,0,0,0],
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 [0,0,0,0,0,0,0,1,1,1,0,0,0],
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 [0,1,1,0,1,0,0,0,0,0,0,0,0],
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 [0,1,0,0,1,1,0,0,1,0,1,0,0],
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 [0,1,0,0,1,1,0,0,1,1,1,0,0],
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 [0,0,0,0,0,0,0,0,0,0,1,0,0],
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 [0,0,0,0,0,0,0,1,1,1,0,0,0],
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 [0,0,0,0,0,0,0,1,1,0,0,0,0]]
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Given the above grid, return6. Note the answer is not 11, because the island must be connected 4-directionally.
Example 2:
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[[0,0,0,0,0,0,0,0]]
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Given the above grid, return0.
Note:The length of each dimension in the givengriddoes not exceed 50.
2. Implementation
(1) DFS
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class Solution {
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    public int maxAreaOfIsland(int[][] grid) {
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        if (grid == null || grid.length == 0) {
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            return 0;
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        }
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​
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        int m = grid.length, n = grid[0].length;
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        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
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        int maxArea = 0, area = 0;
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​
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        for (int i = 0; i < m; i++) {
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            for (int j = 0; j < n; j++) {
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                if (grid[i][j] == 1) {
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                    area = getIslandArea(i, j, grid, directions);
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                    maxArea = Math.max(maxArea, area);
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                }
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            }
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        }
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        return maxArea;
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    }
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​
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    public int getIslandArea(int row, int col, int[][] grid, int[][] directions) {
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        int area = 1;
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        grid[row][col] = 0;
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​
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        for (int[] direction : directions) {
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            int nextRow = row + direction[0];
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            int nextCol = col + direction[1];
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​
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            if (isValid(nextRow, nextCol, grid, directions)) {
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                area += getIslandArea(nextRow, nextCol, grid, directions);
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            }
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        }
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        return area;
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    }
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​
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    public boolean isValid(int row, int col, int[][] grid, int[][] directions) {
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        return row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && grid[row][col] == 1;
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    }
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}
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(2) Union Find
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class Solution {
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    public int maxAreaOfIsland(int[][] grid) {
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        if (grid == null || grid.length == 0) {
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            return 0;
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        }
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​
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        int m = grid.length, n = grid[0].length;
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        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
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​
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        UnionFind uf = new UnionFind(grid);
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​
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        for (int i = 0; i < m; i++) {
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            for (int j = 0; j < n; j++) {
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                if (grid[i][j] == 1) {
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                    for (int[] direction : directions) {
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                        int nextRow = i + direction[0];
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                        int nextCol = j + direction[1];
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​
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                        if (isValid(nextRow, nextCol, grid)) {
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                            int id1 = i * n + j;
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                            int id2 = nextRow * n + nextCol;
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                            uf.union(id1, id2);
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                        }
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                    }
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                }
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            }
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        }
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        return uf.maxArea();
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    }
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​
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    public boolean isValid(int row, int col, int[][] grid) {
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        return row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && grid[row][col] == 1;
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    }
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​
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    class UnionFind {
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        int[] sets;
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        int[] size;
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        int m, n, count, maxSize;
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​
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        public UnionFind(int[][] grid) {
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            m = grid.length;
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            n = grid[0].length;
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            sets = new int[m * n];
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            size = new int[m * n];
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            maxSize = 0;
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​
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            for (int i = 0; i < m; i++) {
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                for (int j = 0; j < n; j++) {
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                    if (grid[i][j] == 1) {
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                        int id = i * n + j;
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                        sets[id] = id;
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                        size[id] = 1;
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                        maxSize = 1;
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                        ++count;
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                    }
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                }
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            }
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        }
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​
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        public int find(int node) {
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            while (node != sets[node]) {
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                node = sets[node];
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            }
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            return node;
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        }
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​
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        public void union(int i, int j) {
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            int node1 = find(i);
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            int node2 = find(j);
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​
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            if (node1 == node2) {
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                return;
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            }
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​
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            if (size[node1] < size[node2]) {
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                sets[node1] = node2;
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                size[node2] += size[node1];
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                maxSize = Math.max(maxSize, size[node2]);
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            }
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            else {
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                sets[node2] = node1;
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                size[node1] += size[node2];
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                maxSize = Math.max(maxSize, size[node1]);
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            }
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​
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            --count;
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        }
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​
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        public int maxArea() {
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            return maxSize;
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        }
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    }
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}
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3. Time & Space Complexity
DFS: 时间复杂度O(mn), 空间复杂度O(max(m,n))
Union Find: 时间复杂度O(mn * log(mn)), 空间复杂度O(mn)
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Contents
695. Max Area of Island
1. Question
2. Implementation
3. Time & Space Complexity
695. Max Area of Island​

1. Question

2. Implementation

3. Time & Space Complexity

695. Max Area of Island
