标签:search leetcode first i++ length bool call sea for
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
1 public class Solution { 2 public int NumIslands(char[,] grid) { 3 if(grid==null||grid.GetLength(0)==0||grid.GetLength(1)==0) 4 { 5 return 0; 6 } 7 int h = grid.GetLength(0); 8 int w = grid.GetLength(1); 9 int count = 0; 10 // Create 2D bool array to record the islands have been visited 11 bool[,] visit = new bool[h,w]; 12 //find the first ‘1‘ on island and find all other ‘1‘s by calling Bfs; 13 for(int i=0; i<h; i++) 14 { 15 for(int j=0; j<w; j++) 16 { 17 //only check unvisited ones 18 if(!visit[i,j] && grid[i,j]==‘1‘) 19 { 20 count++; 21 Bfs(grid, visit , i , j); 22 } 23 } 24 } 25 return count; 26 } 27 //Breadth-first Search to find all ‘1‘s on this island and flip visit to true; 28 private void Bfs(char[,] grid, bool[,] visit, int row, int col) 29 { 30 int h = grid.GetLength(0); 31 int w = grid.GetLength(1); 32 if(row>=0&& row<h && col>=0 && col<w && !visit[row,col] && grid[row,col]==‘1‘) 33 { 34 //flip visit 35 visit[row,col]=true; 36 //element above 37 Bfs(grid, visit, row-1, col); 38 //element below 39 Bfs(grid, visit, row+1, col); 40 //element left 41 Bfs(grid, visit, row, col-1); 42 //element right 43 Bfs(grid, visit, row, col+1); 44 } 45 } 46 }
Leetcode 200. Number of Islands
标签:search leetcode first i++ length bool call sea for
原文地址:http://www.cnblogs.com/MiaBlog/p/6041173.html