LeetCode – Number of Islands (Java)
Given a 2d 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==0grid[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<0i>=mj<0j>=ngrid[i][j]!='1') return; grid[i][j]='X'; merge(grid, i1, j); merge(grid, i+1, j); merge(grid, i, j1); merge(grid, i, j+1); } 
Java Solution 2  UnionFind
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.
<pre><code> String foo = "bar"; </code></pre>

Sudhir

foo

DachiCoding

Guus