题目

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]
]
输出：1

示例 2：

输入：grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]
]
输出：3

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] 的值为 '0' 或 '1'

解答

源代码

class Solution {public int numIslands(char[][] grid) {int count = 0;for (int i = 0; i < grid.length; i++) {for (int j = 0; j <grid[0].length; j++) {if (grid[i][j] == '1') {infect(grid, i, j);count++;}}}return count;}public void infect(char[][] grid, int i, int j) {if (i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == '1') {grid[i][j] = '2';} else {return;}infect(grid, i - 1, j);infect(grid, i + 1, j);infect(grid, i, j - 1);infect(grid, i, j + 1);}
}

总结

这道题给我拓展了一个新的解题思路，讨论区大佬起名为“感染函数”，遍历检测到1时就进入感染函数将这个1能连接到的所有1（包括它本身）都变为2，这样：0代表水，1代表未遍历过的陆地，2代表已经遍历过的陆地。那么就可以做到当到达一个未曾来到过的岛屿时，我们把这片岛屿的陆地全部遍历一遍，那么之后就算再次来到这片岛屿的陆地，也不会错把岛屿数量增加。