200. 岛屿数量

深度优先遍历与广度优先遍历都可

只需要挨个遍历每个点，当且仅该点为‘1’且未被访问时，深度访问该点并将岛屿的数量+1，并且深度遍历与该点"陆地"相连接的所有点(陆地)

/ @lc app=leetcode.cn id=200 lang=cpp [200] 岛屿数量*/// @lc code=start
class Solution {
public:int numIslands(vector<vector<char>>& grid) {int m = grid.size(); int n = grid[0].size();int count = 0; // 用于统计岛屿的数量vector<vector<int>> path(m,vector<int>(n,0)); //用于标记已访问过的陆地for(int i=0;i<m;i++){for(int j=0;j<n;j++){if(grid[i][j] == '1'&&path[i][j] == 0){deep(grid,path,i,j,m,n);path[i][j] = 1;count++;}}}return count;}void deep(vector<vector<char>>& grid,vector<vector<int>>& path,int x,int y,int &m,int &n){if(x>0&&grid[x-1][y]=='1'&&path[x-1][y] == 0){path[x-1][y] = 1;deep(grid,path,x-1,y,m,n);}if(x<m-1&&grid[x+1][y]=='1'&&path[x+1][y] == 0){path[x+1][y] = 1;deep(grid,path,x+1,y,m,n);}if(y>0&&grid[x][y-1]=='1'&&path[x][y-1] == 0){path[x][y-1] = 1;deep(grid,path,x,y-1,m,n);}if(y<n-1&&grid[x][y+1]=='1'&&path[x][y+1] == 0){path[x][y+1] = 1;deep(grid,path,x,y+1,m,n);}}
};
// @lc code=end