岛屿数量

岛屿数量

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出：1
示例 2：

输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/number-of-islands
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解题思路

遍历矩阵中的所有点，对于岛屿使用广度优先搜索or深度优先搜索or并查集找到一个小岛。

code 广度优先搜索

class Solution {
public:
    typedef pair<int,int> PII;
    int dx[4] = {-1,1,0,0};
    int dy[4] = {0,0,-1,1};

    int numIslands(vector<vector<char>>& grid) {

       int m = grid.size(),n = grid[0].size();

       vector<vector<int>> visit(m,vector<int>(n,0));

       int ans = 0;
       for(int i = 0;i < m;i ++)
       {
           for(int j = 0;j < n;j++)
           {
               if(grid[i][j] == '1' && !visit[i][j])
               {
                   ans ++;
                   queue<PII> q;
                   q.push({i,j});
				  //入队表明已经走过
                   visit[i][j] = 1;
                   while(!q.empty())
                   {
                       PII cur = q.front();
                       q.pop();
                       for(int k = 0;k < 4;k++)
                       {
                           int x = cur.first + dx[k];
                           int y = cur.second + dy[k];
                           if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1' && !visit[x][y])
                           {
                               visit[x][y] = 1;
                               q.push({x,y});
                           }
                       }
                   }
               }
           }
       } 

       return ans;
    }
};