leetcode Binary Tree Breadth-First Search

leetcode 200题

Posted by franki on September 6, 2021

leetcode 200题 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

思路

dfs

代码

 // 注意题意 连续大陆就是一个岛屿
 // 遍历二维数组，每当遇到1开启搜索模式，从当前节点向左/右/上/下，每次分别移动一步，如果是1则替换为0
var numIslands = function(grid) {
  function dfs(grid,i,j){
    // 递归终止条件
    if(i<0||i>=grid.length||j<0||j>=grid[0].length||grid[i][j]==='0'){
       return  
    }
    grid[i][j]='0' // 走过的标记为0
    dfs(grid, i + 1, j)
    dfs(grid, i, j + 1)
    dfs(grid, i - 1, j)
    dfs(grid, i, j - 1)
  }
  let count=0
  for(let i=0;i<grid.length;i++){
      for(let j=0;j<grid[0].length;j++){
          if(grid[i][j]==='1'){
              dfs(grid,i,j)
              count++
          }
      }
  }
 return count
};

复杂度分析

  • 时间复杂度：O(N^2)
  • 空间复杂度：O(1)
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