宽度优先搜索（BFS）之习题分析

一、宽度优先搜索的概念

​ 广度优先搜索（也称宽度优先搜索，缩写BFS，以下采用广度来描述）是连通图的一种遍历策略。因为它的思想是从一个顶点V0开始，辐射状地优先遍历其周围较广的区域，故得名。

​ 一般可以用它做什么呢？一个最直观经典的例子就是走迷宫，我们从起点开始，找出到终点的最短路程，很多最短路径算法就是基于广度优先的思想成立的。

二、小岛问题

（一）、题目需求

​ 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'

（二）、解法

public int numIslands(char[][] grid) {
   
    if (grid == null || grid.length == 0) {
   
        return 0;
    }
    if (grid[0] == null || grid[0].length == 0) {
   
        return 0;
    }

    int number = 0;
    int row = grid.length;
    int column = grid[0].length;
    boolean[][] visited = new boolean[row][column];

    for (int i = 0; i < row; i++) {
   
        for (int j = 0; j < column; j++) {
   
            if (grid[i][j] == '1' && !visited[i][j]) {
   
                bfs(grid, i, j, visited);
                number++;
            }
        }
    }
    return number;
}

private void bfs(char[][] grid, int i, int j, boolean[][] visited) {
   
    int[] kx = {
   1, -1, 0, 0};
    int[] ky = {
   0, 0, 1, -1};
    Queue<Integer> xQueue = new LinkedList<>();
    Queue<Integer> yQueue = new LinkedList<>();

    visited[i]