LeetCode – Number of Islands (Java)

Given a 2-d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Java Solution 1 - DFS

The basic idea of the following solution is merging adjacent lands, and the merging should be done recursively.

Each element is visited once only. So time is O(m*n).

public int numIslands(char[][] grid) {
    if(grid==null || grid.length==0||grid[0].length==0)
        return 0;
 
    int m = grid.length;
    int n = grid[0].length;
 
    int count=0;
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
            if(grid[i][j]=='1'){
                count++;
                merge(grid, i, j);
            }
        }
    }
 
    return count;
}
 
public void merge(char[][] grid, int i, int j){
    int m=grid.length;
    int n=grid[0].length;
 
    if(i<0||i>=m||j<0||j>=n||grid[i][j]!='1')
        return;
 
    grid[i][j]='X';
 
    merge(grid, i-1, j);
    merge(grid, i+1, j);
    merge(grid, i, j-1);
    merge(grid, i, j+1);
}

Java Solution 2 - Union-Find

Time is O(m*n*log(k)).

public int numIslands(char[][] grid) {
    if(grid==null || grid.length==0 || grid[0].length==0)
        return 0;
 
    int m = grid.length;
    int n = grid[0].length;
 
    int[] dx={-1, 1, 0, 0};
    int[] dy={0, 0, -1, 1};
 
    int[] root = new int[m*n];
 
    int count=0;
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
            if(grid[i][j]=='1'){
                root[i*n+j] = i*n+j;            
                count++;
            }
        }
    }
 
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
            if(grid[i][j]=='1'){
                for(int k=0; k<4; k++){
                    int x = i+dx[k];
                    int y = j+dy[k];
 
                    if(x>=0&&x<m&&y>=0&&y<n&&grid[x][y]=='1'){
                        int cRoot = getRoot(root, i*n+j);
                        int nRoot = getRoot(root, x*n+y);
                        if(nRoot!=cRoot){
                            root[cRoot]=nRoot; //update previous node's root to be current
                            count--;
                        }
 
                    }
                }
            }
        }
    }
 
    return count;
}
 
public int getRoot(int[] arr , int i){
    while(arr[i]!=i){
        i = arr[arr[i]];
    }
 
    return i;
}

Check out Number of Island II.

Category >> Algorithms >> Interview  
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  • Sudhir

    isn’t the time complexity of DFS O(m)+O(n) where m is number of vertices and n is number of edges.??

  • foo

    def num_islands():

    visited = set()

    islands = 0

    for i in range(4):

    for j in range(5):

    if m[i][j] == 1:

    visited.add((i,j))

    if i == 0 and j == 0:

    islands += 1

    elif i == 0:

    if m[i][j-1] == 1:

    continue # we have visited this island

    else:

    islands += 1

    else:

    if (i-1,j-1) in visited or

    (i-1,j+1) in visited or

    (i-1,j) in visited or

    (i,j-1) in visited:

    continue

    else:

    islands += 1

    print visited

    return islands

  • DachiCoding

    It’s necessary to call merge(gird,i-1,j) and merge(grid,i,j-1).

    e.g
    11111
    00100
    00101 <- this guy cannot be achieved by merge(grid,i+1,j) or merge(grid,i,j+1)
    00111

  • Guus

    You don’t need to call merge(grid, i - 1, j) and merge(grid, i, j - 1).