python 学习 leetcode ---number of island

最新推荐文章于 2025-08-16 17:02:07 发布

转载最新推荐文章于 2025-08-16 17:02:07 发布 · 70 阅读

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原文链接：http://www.cnblogs.com/fanhaha/p/7846788.html

文章标签：

#python #数据结构与算法

本文介绍了一个基于深度优先搜索（DFS）算法的岛屿数量计算方法。该算法通过遍历二维网格地图，利用四叉树结构来识别由'1'（陆地）组成的岛屿，并计算岛屿总数。文章提供了一个Python实现示例。

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Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Answer: 1

Example 2:

Answer: 3

思路 = dfs（四叉树，上下左右） +剪枝

剪枝是剪掉遍历过的结点

class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        count=0
        m=len(grid)
        if m==0:
            return  0
        n=len(grid[0])
        #visit=[[False]*n]*m
        visit = [[False for i in range(n)]for j in range(m)]  
        def check(x, y):  
            if x >= 0 and x<m and y>= 0 and y< n and grid[x][y] == '1' and visit[x][y] == False:  
                return True  
        def dfs(i ,j):
            nbrow = [1,0,-1,0]  
            nbcol = [0,1,0,-1] 
            for d in range(4):
                ni=i+nbrow[d]
                nj=j+nbcol[d]
                if(check(ni,nj)==True):
                    visit[ni][nj]=True
                    dfs(ni,nj)
        for i in range(m):
            for j in range(n):
                if(check(i,j)==True):
                    visit[i][j]=True
                    dfs(i,j)
                    count+=1
        return count

转载于:https://www.cnblogs.com/fanhaha/p/7846788.html