leetcode 200. Number of Islands(DFS)

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原文链接：http://www.cnblogs.com/zywscq/p/5429124.html

本文介绍了一个使用深度优先搜索(DFS)算法解决二维网格中岛屿数量计算的问题。通过遍历每个单元格，当遇到陆地('1')且未被访问过时，则进行一次DFS搜索并计数加一，最终返回岛屿总数。

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

简单的DFS求联通块。

class Solution {
public:
    void dfs(int x,int y,vector<vector<char> >& g){
        used[x][y]=1;
        for(int i=0;i<4;i++){
            int tx=x+dx[i];
            int ty=y+dy[i];
            if(tx>=0&&tx<n&&ty>=0&&ty<m&&used[tx][ty]==0&&g[tx][ty]=='1'){
                dfs(tx,ty,g);
            }
        }
        return;
    }
    int numIslands(vector<vector<char> >& grid) {
        n = grid.size();
        if(n==0) return 0;
        m = grid[0].size();
        used = vector<vector<int> >(n,vector<int>(m,0));
        int ans=0;
        
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(used[i][j]==0&&grid[i][j]=='1'){
                    dfs(i,j,grid);
                    ans++;
                }
            }
        }
        return ans;
    }
    
private:
    int n,m;
    int dx[4]={0,0,1,-1};
    int dy[4]={1,-1,0,0};
    vector<vector<int> >used;
};