朋友圈那題如果有union-find做的話：
```
class Solution:
    def findCircleNum(self, M: List[List[int]]) -> int:
        if not M:
            return
        row, col = len(M), len(M[0])
        self.count = len(M)
        friends = [i for i in range(row)]
        # find
        def find(x):
            if friends[x] != x:
                return find(friends[x])
            return friends[x]
        # unio
        def union(x, y):
            xroot, yroot = find(x), find(y)
            if xroot == yroot:
                return
            friends[xroot] = yroot
            self.count -= 1

        for i in range(row):
            for j in range(col):
                # handle only half the chart
                if i < j and M[i][j] == 1:
                    union(i, j)
        return self.count

```

func numIslands(_ g: [[Character]]) -> Int {
        
        var grid = g
        var count = 0
        
        if grid.count == 0 {
            return 0
        }
    
        
        let Rows = grid.count
        let Cols = grid.first?.count ?? 0
        
        
        for i in 0..<Rows {
            for j in 0..<Cols {
                if grid[i][j] == "1" {
                    count += 1
                    renderOneToZero(&grid,Rows: Rows,Cols: Cols,i: i,j: j)
                }
            }
        }
        
        return count
    }
    
    func renderOneToZero(_ grid: inout [[Character]], Rows:Int,Cols:Int, i:Int, j:Int) {
        
        if i >= Rows || j >= Cols || i < 0 || j < 0 {
            return
        }
        
        if grid[i][j] == "1" {
            grid[i][j] = "0"
            //把 上 左 下 右
            //左
            renderOneToZero(&grid, Rows: Rows, Cols: Cols, i: i, j: j-1)
            //右
            renderOneToZero(&grid, Rows: Rows, Cols: Cols, i: i , j: j + 1)
            //上
            renderOneToZero(&grid, Rows: Rows, Cols: Cols, i: i - 1 , j: j)
            //下面
            renderOneToZero(&grid, Rows: Rows, Cols: Cols, i: i + 1 , j: j)
        }
    }