岛屿数量

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#岛屿数量

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本文介绍了一种使用广度优先搜索（BFS）和深度优先搜索（DFS）算法来计算二维网格中岛屿数量的方法。通过遍历网格，每当遇到岛屿时，使用BFS或DFS进行搜索并将已访问的部分标记为已处理，以此来准确计算岛屿的总数。

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BFS方法，每次发现为‘1’的岛屿才搜索，且只要搜到连通域就将其置‘0’：

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

        len_i = len(grid)
        if len_i == 0:
            return 0
        len_j = len(grid[0])

        numLands = 0

        from collections import deque

        for m in range(len_i): #以这种方式对每个位置搜索
            for n in range(len_j):
                if grid[m][n] == '1':
                    numLands += 1
                    stack = deque([[m, n]])
                    grid[m][n] = '0'
                    while stack:
                        i, j = stack.popleft()
                        if i + 1 < len_i and grid[i + 1][j] == '1':
                            stack.append([i + 1, j])
                            grid[i + 1][j] = '0' #搜索到就置0

                        if i - 1 >= 0 and grid[i - 1][j] == '1':
                            stack.append([i - 1, j])
                            grid[i - 1][j] = '0'

                        if j + 1 < len_j and grid[i][j + 1] == '1':
                            stack.append([i, j + 1])
                            grid[i][j + 1] = '0'

                        if j - 1 >= 0 and grid[i][j - 1] == '1':
                            stack.append([i, j - 1])
                            grid[i][j - 1] = '0'
        return numLands

dfs解法：

from collections import deque

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        
        def __dfs(grid, len_i, len_j, pos):
            i, j = pos[0], pos[1]
            directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
            for di, dj in directions:
                if 0 <= (i + di) < len_i and 0 <= (j + dj) < len_j and grid[i + di][j + dj] == '1':
                    grid[i + di][j + dj] = '0'
                    __dfs(grid, len_i, len_j, [i + di, j + dj])

        len_i = len(grid)
        if len_i == 0:
            return 0
        len_j = len(grid[0])

        numLands = 0

        for m in range(len_i):
            for n in range(len_j):
                if grid[m][n] == '1':
                    numLands += 1
                    grid[m][n] = '0'
                    __dfs(grid, len_i, len_j, (m, n))
                    
        return numLands