【CSDN 每日一练 ★★☆】【DSF/BSF】岛屿数量

Jim-2ha02023-10-31 19:08

【CSDN 每日一练 ★★☆】【DSF/BSF】岛屿数量

深度遍历 宽度遍历

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

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'

Java实现

java 复制代码

public int numIslands(char[][] grid) {
    int islandNum = 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); //将与grid[i][j]相连的`1`标记为`2`
                islandNum++;
            }
        }
    }
    return islandNum;
}
public void infect(char[][] grid, int i, int j) {
    // 如果发现已经被标记或者是水，结束深度遍历
    if (i < 0 || i >= grid.length ||
            j < 0 || j >= grid[0].length || grid[i][j] != '1') {
        return;
    }
    grid[i][j] = '2';
    // 宽度遍历四周
    infect(grid, i + 1, j);
    infect(grid, i - 1, j);
    infect(grid, i, j + 1);
    infect(grid, i, j - 1);
}