littlearAppearance
题目描述
200.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'
思路
两层循环遍历网格，外层循环x,内层循环y，如果grid[x][y]是陆地，且未被访问过(定义一个visied数组)，则递归判断上下左右4个方向是不是也是陆地
当x,y越界(x<0 || x>=m || y<0 || y>=n)、grid[x][y]不是陆地且被访问过，就可以提前退出递归
题解
javascript
var numIslands = function(grid) {
let m = grid.length,n=grid[0].length,count=0,
visied=Array.from({length: m}, ()=>Array(n).fill(false)),
dirs = [[-1,0],[0,1],[1,0],[0,-1]];
let dfs = (x, y) => {
if(x<0 || x>=m || y<0 || y>=n || grid[x][y]==='0' || visied[x][y]) return;
visied[x][y]=true;
for(let [dx, dy] of dirs){
dfs(x+dx,y+dy)
}
return
}
for(let x=0;x<m;x++){
for(let y=0;y<n;y++){
if(grid[x][y]==1 && !visied[x][y]){
count++;
dfs(x,y);
}
}
}
return count
}