200.岛屿数量

链接

思路 ： 遍历二维矩阵，碰到1就岛屿数量+1 然后通过floodfill，将于遍历到1点陆地想连的陆地全部清除(变为""0)

代码

/**
* @param {character[][]} grid
* @return {number}
*/
var numIslands = function(grid) {
   let [n,m] = [grid.length,grid[0].length]
   let count = 0;
   for(let i = 0;i<n;i++){
       for(let j = 0;j<m;j++){
           if(grid[i][j] == "1"){
               count+=1;
               dfs(grid,i,j)
           }
       }
   }
   return count
};

let dfs = (grid,i,j) =>{
   let [n,m] = [grid.length,grid[0].length]
   if(i<0||j<0||i>=n||j>=m){
       return
   }
   if(grid[i][j] == "0"){
       return
   }
   grid[i][j] = "0"
   dfs(grid,i-1,j)
   dfs(grid,i+1,j)
   dfs(grid,i,j-1)
   dfs(grid,i,j+1)
}