Leetcode: Number of Islands

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



Example 2:



11000

11000

00100

00011

Answer: 3

DFS的Flood Fill方法，

使用额外Visited数组的做法：

 1 public class Solution {

 2     public int numIslands(char[][] grid) {

 3         if (grid==null || grid.length==0 || grid[0].length==0) return 0;

 4         int count = 0;

 5         boolean[][] visited = new boolean[grid.length][grid[0].length];

 6         for (int i=0; i<grid.length; i++) {

 7             for (int j=0; j<grid[0].length; j++) {

 8                 if (grid[i][j] != '1') continue;

 9                 else {

10                     count++;

11                     floodFill(grid, i, j, visited);

12                 }

13             }

14         }

15         return count;

16     }

17     

18     public void floodFill(char[][] grid, int i, int j, boolean[][] visited) {

19         if (i<0 || i>=grid.length || j<0 || j>=grid[0].length) return;

20         if (visited[i][j]) return;

21         if (grid[i][j] != '1') return;

22         grid[i][j] = '2';

23         floodFill(grid, i-1, j, visited);

24         floodFill(grid, i+1, j, visited);

25         floodFill(grid, i, j-1, visited);

26         floodFill(grid, i, j+1, visited);

27     }

28 }

更节省空间的方法：不使用额外visited数组，但是用‘1’变成‘2’表示visited的方法

 1 public class Solution {

 2     public int numIslands(char[][] grid) {

 3         if (grid==null || grid.length==0 || grid[0].length==0) return 0;

 4         int count = 0;

 5         for (int i=0; i<grid.length; i++) {

 6             for (int j=0; j<grid[0].length; j++) {

 7                 if (grid[i][j] != '1') continue;

 8                 else {

 9                     count++;

10                     floodFill(grid, i, j);

11                 }

12             }

13         }

14         return count;

15     }

16     

17     public void floodFill(char[][] grid, int i, int j) {

18         if (i<0 || i>=grid.length || j<0 || j>=grid[0].length) return;

19         if (grid[i][j] != '1') return; //either 0(water) or 2(visited)

20         grid[i][j] = '2';

21         floodFill(grid, i-1, j);

22         floodFill(grid, i+1, j);

23         floodFill(grid, i, j-1);

24         floodFill(grid, i, j+1);

25     }

26 }