前言

记录下leetcode每日一题

题目描述

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

根据题解，1为陆地，0为海水，一块陆地左右上下相连的部分被认为一个岛屿。从二维数组第一项开始，如果是1，则对岛屿数量+1，并对其相连的进行深度优先遍历，每一次扫描，如果没有超过边界,将其标记为0防止重复遍历

代码

/**
 * @param {character[][]} grid
 * @return {number}
 */
var numIslands = function (grid) {
    let col = grid[0].length;// 列
    let row = grid.length // 行 
    let num = 0
    for (let i = 0; i < row; i++) {
        for (let j = 0; j < col; j++) {
            if (grid[i][j] === '1') {
                num++
                // 开启深度优先
                dfs(grid, i, j)
            }
        }
    }
    return num
};

function dfs(grid, row, col) {
    // 边界判断
    if (row < 0 || row >= grid.length || col < 0 || col >= grid[0].length || grid[row][col] === '0') return false
    grid[row][col] = '0' // 将状态变为0 表示已经搜索过
    // 遍历上下左右
    // 遍历上面
    dfs(grid, row - 1, col)
    // 遍历下面
    dfs(grid, row + 1, col)
    // 遍历左边
    dfs(grid, row, col - 1)
    // 遍历右边
    dfs(grid, row, col + 1)
}

最后

每天进步一点点