【LeetCode】200.岛屿数量

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 = 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。