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:

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.

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Guus on 2015-11-7

You don’t need to call merge(grid, i - 1, j) and merge(grid, i, j - 1).
DachiCoding on 2016-1-13

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
foo on 2016-4-4

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
Sudhir on 2016-9-8

isn’t the time complexity of DFS O(m)+O(n) where m is number of vertices and n is number of edges.??
Misha Arora on 2017-2-24

In the second method, complexity is m*n*log(k). What is k??

LeetCode – Number of Islands (Java)

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