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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
思路：
leetcode的新题，题目思路很简单，DFS或者BFS，而且还不用回溯，直接判断，难度不大。因为自己用弱项BFS写出来了，贴出来备忘。
题解：
class Solution { public: int dir[4][2] = {{-1,0}, {0,-1}, {1,0},{0,1}}; void dfs(vector<vector<char> > &grid, int x, int y) { grid[x][y] = ‘0‘; for(int k=0;k<4;k++) { int nx = x+dir[k][0]; int ny = y+dir[k][1]; if(nx>=0 && nx<grid.size() && ny>=0 && ny<grid[0].size()) if(grid[nx][ny]==‘1‘) dfs(grid, nx, ny); } } int numIslands(vector<vector<char> > &grid) { int m = grid.size(); if(m==0) return 0; int n = grid[0].size(); int res = 0; for(int i=0;i<m;i++) for(int j=0;j<n;j++) if(grid[i][j]==‘1‘) { dfs(grid, i, j); res++; } return res; } };
class Solution { public: struct point { point(int xx=0, int yy=0): x(xx), y(yy) {} int x; int y; }; int dir[4][2] = {{0,-1}, {-1,0}, {0,1}, {1,0}}; void BFS(vector<vector<char> > &grid, int x, int y) { queue<point> q; int curx, cury; q.push(point(x,y)); while(!q.empty()) { curx = q.front().x; cury = q.front().y; q.pop(); grid[curx][cury] = ‘0‘; for(int k=0;k<4;k++) { int nx = curx+dir[k][0]; int ny = cury+dir[k][1]; if(nx>=0 && nx<grid.size() && ny>=0 && ny<grid[0].size()) if(grid[nx][ny]==‘1‘) { q.push(point(nx,ny)); grid[nx][ny] = ‘0‘; } } } } int numIslands(vector<vector<char> > &grid) { int m = grid.size(); if(m==0) return 0; int n = grid[0].size(); int res = 0; for(int i=0;i<m;i++) for(int j=0;j<n;j++) { if(grid[i][j] == ‘1‘) { ++res; BFS(grid, i,j); } } return res; } };
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原文地址：http://www.cnblogs.com/jiasaidongqi/p/4420914.html
