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:

Answer: 1

Example 2:

Answer: 3

Analyse: BFS or DFS. If found a '1', label that element as visited, and find all elements could be reached from that element and label them.

Runtime: Iteration 12ms

 1 class Solution {
 2 public:
 3     int numIslands(vector<vector<char>>& grid) {
 4         if(grid.empty() || grid[0].empty()) return 0;
 5         
 6         int result = 0;
 7         for(int i = 0; i < grid.size(); i++) {
 8             for(int j = 0; j < grid[i].size(); j++) {
 9                 if(grid[i][j] == '1') {
10                     grid[i][j] = '2';
11                     bfs(grid, i, j);
12                     result++;
13                 }
14             }
15         }
16         return result;
17     }
18     
19     void bfs(vector<vector<char> >& grid, int i, int j) {
20         queue<pair<int, int> > qu;
21         qu.push(make_pair(i, j));
22         
23         int m = grid.size(), n = grid[0].size();
24         while(!qu.empty()) {
25             pair<int, int> node = qu.front();
26             i = node.first;
27             j = node.second;
28             qu.pop();
29             // check whether its neighbour is island
30             if(i > 0 && grid[i - 1][j] == '') {
31                 qu.push(make_pair(i - 1, j));
32                 grid[i - 1][j] = '2';
33             }
34             if(j > 0 && grid[i][j - 1] == '1') {
35                 qu.push(make_pair(i, j - 1));
36                 grid[i][j - 1] = '2';
37             }
38             if(j < n - 1 && grid[i][j + 1] == '1') {
39                 qu.push(make_pair(i, j + 1));
40                 grid[i][j + 1] = '2';
41             }
42             if(i < m - 1 && grid[i + 1][j] == '1') {
43                 qu.push(make_pair(i + 1, j));
44                 grid[i + 1][j] = '2';
45             }
46         }
47     }
48 };

View Code

Runtime: Recursive 8ms.

 1 class Solution {
 2 public:
 3     int numIslands(vector<vector<char>>& grid) {
 4         int result;
 5         if(grid.empty() || grid[0].empty()) return result;
 6         
 7         int row = grid.size(), col = grid[0].size();
 8         for(int i = 0; i < row; i++){
 9             for(int j = 0; j < col; j++){
10                 if(grid[i][j] == '1'){
11                     helper(grid, i, j);
12                     result++;
13                 }
14             }
15         }
16         return result;
17     }
18     
19     void helper(vector<vector<char> > &grid, int x, int y){
20         if(x >= grid.size() || x < 0 || y >= grid[0].size() || y < 0 || grid[x][y] != '1')
21             return;
22             
23         grid[x][y] = '2';
24         helper(grid, x + 1, y);
25         helper(grid, x - 1, y);
26         helper(grid, x, y + 1);
27         helper(grid, x, y - 1);
28     }
29 };

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原文地址：https://www.cnblogs.com/amazingzoe/p/5759327.html