給你一個由?'1'（陸地）和 '0'（水）組成的的二維網(wǎng)格，請你計算網(wǎng)格中島嶼的數(shù)量。島嶼總是被水包圍，并且每座島嶼只能由水平方向和/或豎直方向上相鄰的陸地連接形成。此外，你可以假設該網(wǎng)格的四條邊均被水包圍。示例 1：輸入：grid = [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"] ] 輸出：1 示例 2：輸入：grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"] ] 輸出：3提示：m == grid.length n == grid[i].length 1 <= m, n <= 300 grid[i][j] 的值為 '0' 或 '1'鏈接：https://leetcode-cn.com/problems/number-of-islandsclass Solution:def numIslands(self, grid: List[List[str]]) -> int:row, col, count = len(grid), len(grid[0]), 0def dfs(x, y):gris[x][y] = '0'for c in [(0,1),(0,-1),(-1,0),(1,0)]:nx, ny = x + c[0], y + c[1]if 0<= nx < row and 0<= ny < col and grid[nx][ny] == '1':dfs(nx, ny)for i in range(row):for j in range(col):if grid[i][j] == '1':dfs(i,j)count += 1return count

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