leetcode-mid-Linked list- 200. Number of Islands¶

mycode  57.92%

class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        def recursive(i,j,row,col):
            if i>=0 and i<row and j>=0 and j<col:
                if grid[i][j] == '0':
                    return 
                grid[i][j] = '0'
                if i-1 >= 0:
                    recursive(i-1,j,row,col)
                if i+1 < row:
                    recursive(i+1,j,row,col)
                if j-1 >= 0:
                    recursive(i,j-1,row,col)
                if j+1 <= col:
                    recursive(i,j+1,row,col)
            else:
                return 
        if not grid:
            return 0
        row = len(grid)
        col = len(grid[0])
        for i in range(row):
            for j in range(col):
                if grid[i][j] == '1':
                    #print('if..',grid)
                    recursive(i,j,row,col)
                    grid[i][j] = '1'
                    #print('if..',grid)
        count = 0 
        for i in range(row):
            for j in range(col):
                if grid[i][j] == '1':
                    count += 1
        return count

参考：

思路：其实第二次双层for循环是多余的，可以在前面就计数

class Solution:
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        res = 0
        for r in range(len(grid)):
            for c in range(len(grid[0])):
                if grid[r][c] == "1":
                    self.dfs(grid, r, c)
                    res += 1
        return res
        
    def dfs(self, grid, i, j):
        dirs = [[-1, 0], [0, 1], [0, -1], [1, 0]]
        grid[i][j] = "0"
        for dir in dirs:
            nr, nc = i + dir[0], j + dir[1]
            if nr >= 0 and nc >= 0 and nr < len(grid) and nc < len(grid[0]):
                if grid[nr][nc] == "1":
                    self.dfs(grid, nr, nc)