200. Number of Islands Leetcode Python

最新推荐文章于 2021-11-27 12:16:39 发布

hyperbolechi

最新推荐文章于 2021-11-27 12:16:39 发布

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分类专栏： leetcode 文章标签： leetcode DFS

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本文详细解释了如何使用深度优先搜索算法来计算二维网格中岛屿的数量。通过遍历网格并标记已访问的位置，作者展示了如何准确地计算出给定地图上的岛屿总数。

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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

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

We can do DFS in this problem and mark the position as visited everytime. out of the DFS count the island.

class Solution:
    # @param grid, a list of list of characters
    # @return an integer
    def numIslands(self, grid):
        m = len(grid)
        if m == 0:
            return 0
        n = len(grid[0])
        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(x,y):
            nbrow = [1,0,-1,0]
            nbcol = [0,1,0,-1]
            for k in range(4):
                newx = x + nbrow[k]
                newy = y + nbcol[k]
                if check(newx, newy):
                    visit[newx][newy] = True
                    dfs(newx,newy)
        count = 0
        for row in range(m):
            for col in range(n):
                if check(row,col):
                    visit[row][col] = True
                    dfs(row,col)
                    count+=1
        return count