Python 岛屿问题

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原创最新推荐文章于 2026-09-16 14:06:12 发布 1.6k 阅读 AI 写同款· GEO 优化 ›

孤岛问题（Number of Islands） python解法 + 思路给定一个由 '1'（陆地）和 '0'（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。示例 1: 输入: 11110 11010 11000 00000 输出: 1 示例 2: 输入: 11000 11000 00100 00011 输出: 3 由于在今年3月份网上笔试的时候碰到过这... 阅读详情

import numpy as np
import pandas as pd
# from pydantic
nums = [
  [1, 1, 0, 0, 0],
  [0, 1, 0, 0, 1],
  [0, 0, 0, 1, 1],
  [0, 0, 0, 0, 0],
  [0, 0, 0, 0, 1]
]

class Solution(object):
    def numIslands(self, grid):
        row = len(grid)
        if row == 0 or grid == None:
            return 0
        col = len(grid[0])
        count = 0
        for i in range(row):
            for j in range(col):
                if grid[i][j] == 1:
                    self.dfsFind(grid, i, j, row, col)
                    count += 1
                    print(count)
        return count
    def dfsFind(self, grid, i, j, row, col):
        print(i,j,row,col)
        #   [1, 1, 0, 0, 0],
        #   [0, 1, 0, 0, 1],
        #   [0, 0, 0, 1, 1],
        #   [0, 0, 0, 0, 0],
        #   [0, 0, 0, 0, 1]
        if grid[i][j] == 1:
            grid[i][j] = 0
            if i - 1 >= 0:
                grid = self.dfsFind(grid, i - 1, j, row, col)
            if i + 1 < row:
                grid = self.dfsFind(grid, i + 1, j, row, col)
            if j - 1 >= 0:
                grid = self.dfsFind(grid, i, j - 1, row, col)
            if j + 1 < col:
                grid = self.dfsFind(grid, i, j + 1, row, col)
        return grid

a=Solution()
a.numIslands(nums)