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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
解题思路
将二维矩阵看作图，采用广度优先遍历
扫描矩阵，遇到陆地则将其加入队列，
当队列非空是，弹出队首，然后以它为基础将四个方向中为陆地的加入队列
元素加入队列时要将该元素设为"水"，防止重复遍历
广度优先遍历的次数就是岛屿数量
代码
class Solution {
int rows, cols;
public int numIslands(char[][] grid) {
rows = grid.length;
cols = grid[0].length;
int count = 0;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (grid[i][j] == '1') {
count++;
grid[i][j] = '0';
Deque<Integer> deque = new LinkedList<>();
deque.addLast(i * cols + j); // 二维坐标转为一维索引
while (!deque.isEmpty()) {
int index = deque.removeFirst();
int row = index / cols, col = index % cols;
helper(grid, row + 1, col, deque);
helper(grid, row - 1, col, deque);
helper(grid, row, col + 1, deque);
helper(grid, row, col - 1, deque);
}
}
}
}
return count;
}
private void helper(char[][] grid, int i, int j, Deque<Integer> deque) {
if (i >= 0 && i < rows && j >= 0 && j < cols && grid[i][j] == '1') {
grid[i][j] = '0';
deque.addLast(i * cols + j);
}
}
}
