标签：private sea turn cat ges position empty char vector
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
00000Answer: 1
Example 2:
11000
11000
00100
00011Answer: 3
This problem is typical BFS and DFS, either approach can solve it. The idea is pretty straightforward.
Time complexity is O(m * n). Space complexity is O(1).
class Solution { public: int numIslands(vector<vector<char>>& grid) { if(grid.empty()){ return 0; } int cnt = 0; int row = grid.size(); int col = grid[0].size(); vector<pair<int, int>> dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; for(int i = 0; i < row; i++){ for(int j = 0; j < col; j++){ if(grid[i][j] == ‘1‘){ cnt++; grid[i][j] = ‘0‘; queue<pair<int, int>> que; que.push({i, j}); while(!que.empty()){ pair<int, int> cur = que.front(); que.pop(); for(auto dir : dirs){ int x = cur.first + dir.first; int y = cur.second + dir.second; if(x >= 0 && x < row && y >= 0 && y < col && grid[x][y] == ‘1‘){ que.push({x, y}); grid[x][y] = ‘0‘; } } } } } } return cnt; } };
class Solution { public: int numIslands(vector<vector<char>>& grid) { if(grid.empty()){ return 0; } int cnt = 0; for(int i = 0; i < grid.size(); i++){ for(int j = 0; j < grid[0].size(); j++){ if(grid[i][j] == ‘1‘){ grid[i][j] = ‘0‘; cnt++; island_search(grid, i, j); } } } return cnt; } private: void island_search(vector<vector<char>>& grid, int i, int j){ vector<pair<int,int>> dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; for(auto dir : dirs){ int x = dir.first + i; int y = dir.second + j; if(x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size() && grid[x][y] == ‘1‘){ grid[x][y] = ‘0‘; island_search(grid, x, y); } } return; } };
标签：private sea turn cat ges position empty char vector
原文地址：http://www.cnblogs.com/wdw828/p/6960910.html
