[LeetCode] 200. Number of Islands

접근

• 오히려 easy난이도보다 쉬운 문제인 것 같다
• 제공되는 자료형이 숫자가 아니라 문자임을 감안하지 않아 조금 고민
• 일반적인 방법의 코드도 숙지하는 것이 중요해 보임

 

문제

https://leetcode.com/problems/number-of-islands/description/

 

솔루션

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:

        answer = 0
        ROW_length = len(grid)
        COL_length = len(grid[0])
        visited = [[False for _ in range(COL_length)] for _ in range(ROW_length)]

        def dfs(ROW, COL):            
            # update this point as visited point
            visited[ROW][COL] = True
            
            # check left side
            if COL > 0 and grid[ROW][COL-1] == '1' and \
                visited[ROW][COL-1] == False:
                dfs(ROW, COL-1)

            # check right side
            if COL < COL_length-1 and grid[ROW][COL+1] == '1' and \
                visited[ROW][COL+1] == False:
                dfs(ROW, COL+1)
            
            # check up side
            if ROW > 0 and grid[ROW-1][COL] == '1' and \
                visited[ROW-1][COL] == False:
                dfs(ROW-1, COL)

            # check bottom side
            if ROW < ROW_length-1 and grid[ROW+1][COL] == '1' and \
                visited[ROW+1][COL] == False:
                dfs(ROW+1, COL)
            
        for r in range(ROW_length):
            for c in range(COL_length):
                if visited[r][c] == False and \
                    grid[r][c] == '1':
                    answer += 1
                    dfs(r,c)

        return answer