https://leetcode-cn.com/problems/number-of-islands/
点击查看【bilibili】

题目

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 = [
  2.   ["1","1","1","1","0"],
  3.   ["1","1","0","1","0"],
  4.   ["1","1","0","0","0"],
  5.   ["0","0","0","0","0"]
  6. ]
  7. 输出：1
  1. 输入：grid = [
  2.   ["1","1","0","0","0"],
  3.   ["1","1","0","0","0"],
  4.   ["0","0","1","0","0"],
  5.   ["0","0","0","1","1"]
  6. ]
  7. 输出：3

解答

沉没：如果找到1，把周边的1都变成0（沉没掉），记录小岛+1

从第一个点开始，如果是1，由1改为0，向四周扩散
image.png
判断保证不越界
image.png
依次判断周边的点是否是1
image.png

答案

  1. // bfs-（宽）广度优先搜索
  2. // dfs- 深度优先搜索
  3. var numIslands = function(grid) {
  4.   let count = 0
  6.   function dfs(row,col) {
  7.     // 判断越界和为0的情况
  8.     if(row<0 || row>=grid.length || col<0 || col>=grid[0].length || grid[row][col] === '0') {
  9.       return
  10.     }
  11.     grid[row][col] = '0'
  12.     // 向四周扩散
  13.     dfs(row-1, col)
  14.     dfs(row+1, col)
  15.     dfs(row, col-1)
  16.     dfs(row, col+1)
  17.   }
  19.   for(let row=0;row<grid.length;row++) {
  20.     for(let col=0;col<grid[0].length;col++) {
  21.       if(grid[row][col] === '1') {
  22.         count++;
  23.         dfs(row,col)
  24.       }
  25.     }
  26.   }
  28.   return count
  29. };