Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

深度优先搜索,每次遇到1,则岛的数量+1,从这个1开始找到所有相连的1,将其改为0. public class Solution { private void dfsSearch(char[,] grid, int i, int j, int rows, int cols) { || i >= rows || j < || j >= cols) { return; } ') { return; } grid[i, j] = '; dfsSearch(grid, i + , j, rows…

给定一个由 '1'(陆地)和 '0'(水)组成的的二维网格,计算岛屿的数量.一个岛被水包围,并且它是通过水平方向或垂直方向上相邻的陆地连接而成的.你可以假设网格的四个边均被水包围. 示例 1: 输入:11110110101100000000 输出: 1示例 2: 输入:11000110000010000011 输出: 3 来源:力扣(LeetCode)链接:https://leetcode-cn.com/problems/number-of-islands 方法一:深度优先搜索 [通过]直觉 将…

给定一个由 '1'(陆地)和 '0'(水)组成的的二维网格,计算岛屿的数量.一个岛被水包围,并且它是通过水平方向或垂直方向上相邻的陆地连接而成的.你可以假设网格的四个边均被水包围. 示例 1: 输入: 11110 11010 11000 00000 输出: 1 示例 2: 输入: 11000 11000 00100 00011 输出: 3 广度优先搜索 class Solution { public: int dx[4] = {1, -1, 0, 0}; int dy[4] = {0, 0, 1…

""" Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all…

给定一个由 '1'(陆地)和 '0'(水)组成的的二维网格,计算岛屿的数量.一个岛被水包围,并且它是通过水平方向或垂直方向上相邻的陆地连接而成的.你可以假设网格的四个边均被水包围. 示例 1: 输入: 11110 11010 11000 00000 输出: 1 示例 2: 输入: 11000 11000 00100 00011 输出: 3 //章节 - 队列和栈 //二.队列与广度优先搜索 //1.岛屿的个数 /* 算法思想: 这是一道很有意思的题,本质是求矩阵中连续区域的个数,很容易想到需要用…

并查集算法,也叫Union-Find算法,主要用于解决图论中的动态连通性问题. Union-Find算法类 这里直接给出并查集算法类UnionFind.class,如下: /** * Union-Find 并查集算法 * @author Chiaki */ public class UnionFind { // 连通分量个数 private int count; // 存储若干棵树 private int[] parent; // 记录树的"重量" private int[] size…