Question

我有这个Python代码，用于计算二进制矩阵中的岛数（分组为1）。我如何修改它以考虑顶部和底部的包裹？

class Solution(object):

rowLen = None
colLen = None

def numIslands(self, grid):
    """
    :type grid: List[List[str]]
    :rtype: int
    """

    self.rowLen = len(grid)

    if self.rowLen == 0:
        return 0;

    self.colLen = len(grid[0])

    count = 0;

    for row in range(self.rowLen):
        for col in range(self.colLen):
            if grid[row][col] == '1':
                count += 1
                self.search(grid, row, col)

    return count

def search(self, grid, row, col):
    if (row >= 0 and col >= 0 and row < self.rowLen and col < self.colLen and grid[row][col] == '1'):
        grid[row][col] = 0;

        self.search(grid, row - 1, col)
        self.search(grid, row + 1, col)
        self.search(grid, row, col - 1)
        self.search(grid, row, col + 1)

这当前返回的计数为2，但是应该返回1，因为右下岛应该在左下角用数字包裹后计算为较大岛屿的一部分。

Answer 1

这不明显吗？你的搜索功能只需要环绕而不是在碰到边缘时停止。

def search(self, grid, row, col):
    if row < 0:
        row += self.rowLen
    elif row >= self.rowlen:
        row -= self.rowlen
    if col < 0:
        col += self.colLen
    elif col >= self.colLen:
        col -= self.colLen
    if grid[row][col] == '1':
        grid[row][col] = 0;

        self.search(grid, row - 1, col)
        self.search(grid, row + 1, col)
        self.search(grid, row, col - 1)
        self.search(grid, row, col + 1)

这可以在模运算方面更简单地表达：

def search(self, grid, row, col):
    row %= self.rowlen
    col %= self.colLen
    if grid[row][col] == '1':
        grid[row][col] = 0;

        self.search(grid, row - 1, col)
        self.search(grid, row + 1, col)
        self.search(grid, row, col - 1)
        self.search(grid, row, col + 1)

二进制矩阵中的岛数（包裹）

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