LeetCode算法题200：岛屿的个数解析

给定一个由 ‘1’（陆地）和 ‘0’（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。

示例 1:

输入:
11110
11010
11000
00000

输出: 1

示例 2:

输入:
11000
11000
00100
00011

输出: 3

这个题还是用DFS进行遍历，遇到1且没有遍历过就搜索它周围所有的1。

C++源代码：

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        if(grid.empty() || grid[0].empty()) return 0;
        int m = grid.size(), n = grid[0].size(), res=0;
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='1' && !visited[i][j]){
                    DFS(grid, visited, i, j);
                    res++;
                }
            }
        }
        return res;
    }
    void DFS(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y){
        if(x<0 || x>=grid.size()) return;
        if(y<0 || y>=grid[0].size()) return;
        if(grid[x][y]!='1' || visited[x][y]) return;
        visited[x][y]=true;
        DFS(grid, visited, x-1, y);
        DFS(grid, visited, x+1, y);
        DFS(grid, visited, x, y-1);
        DFS(grid, visited, x, y+1);
    }
};

python3源代码：

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        if len(grid)==0 or len(grid[0])==0:
            return 0
        m = len(grid)
        n = len(grid[0])
        res = 0
        visited = [[0 for i in range(n)] for j in range(m)]
        for i in range(m):
            for j in range(n):
                if grid[i][j]=='1' and visited[i][j]==0:
                    self.DFS(grid, visited, i, j)
                    res += 1
        return res
    def DFS(self, grid, visited, x, y):
        if x<0 or x>=len(grid): return
        if y<0 or y>=len(grid[0]): return
        if grid[x][y]!='1' or visited[x][y]==1: return
        visited[x][y] = 1
        self.DFS(grid, visited, x+1, y)
        self.DFS(grid, visited, x-1, y)
        self.DFS(grid, visited, x, y+1)
        self.DFS(grid, visited, x, y-1)