[抄题]: Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct …
两种方式处理已经访问过的节点:一种是用visited存储已经访问过的1:另一种是通过改变原始数值的值,比如将1改成-1,这样小于等于0的都会停止. Number of Islands 用了第一种方式,Number of Distinct Islands用了第二种方式 200. Number of Islands 时间复杂度o(m*n) 1.这种写法要改变原始输入数组的值 错误版本: 条件判断顺序写错:grid[x][y] == '0' || x < 0 || x >= length || y…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 日期 题目地址:https://leetcode-cn.com/problems/number-of-distinct-islands/ 题目描述 Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) con…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…
Leetcode之深度优先搜索(DFS)专题-200. 岛屿数量(Number of Islands) 深度优先搜索的解题详细介绍,点击 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 分析:这题同样是求连…
题目: 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. 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 lan…
题目如下: Given a 2D grid consists of 0s (land) and 1s (water).  An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s. Return the number of closed islands. E…
题目: 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…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS BFS 日期 题目地址:https://leetcode.com/problems/number-of-islands/description/ 题目描述 Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is…
将所有后缀按照字典序排序后,每新加进来一个后缀,它将产生n - sa[i]个前缀.这里和小罗论文里边有点不太一样. height[i]为和字典序前一个的LCP,所以还要减去,最终累计n - sa[i] - height[i]即可. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; char s[maxn]; int sa[maxn], rank[maxn]…
numbers <- c(4,23,4,23,5,43,54,56,657,67,67,435,         453,435,324,34,456,56,567,65,34,435) a <- table(numbers) a a[names(a)==435] as.data.frame(table(numbers)) sum(numbers == 435)…
地址 https://leetcode-cn.com/contest/weekly-contest-162/problems/number-of-closed-islands/ 有一个二维矩阵 grid ,每个位置要么是陆地(记号为 0 )要么是水域(记号为 1 ). 我们从一块陆地出发,每次可以往上下左右 4 个方向相邻区域走,能走到的所有陆地区域,我们将其称为一座「岛屿」. 如果一座岛屿 完全 由水域包围,即陆地边缘上下左右所有相邻区域都是水域,那么我们将其称为 「封闭岛屿」. 请返回封闭岛…
dfs的第一题 被边界和0包围的1才是岛屿,问题就是分理出连续的1 思路是遍历数组数岛屿,dfs四个方向,遇到1后把周围连续的1置零,代表一个岛屿. /* 思路是:遍历二维数组,遇到1就把周围连续的1变成0,res+1,然后继续遍历,直到结束 周围连续1置零用的是dfs,向四个位置搜索,遇到0返回 */ public int numIslands(char[][] grid) { if (grid.length==0) return 0; int r = grid.length; int c =…
不知道是做着故意放的还是什么原因.总之运行后就会出现问题(奇怪的条目的数量) function merge(left, right){ var result = []; while (left.length > 0 && right.length > 0){ if (left[0] < right[0]){ result.push(left.shift()); } else { result.push(right.shift()); } } return result.c…
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should retu…
在二维空间中有许多球形的气球.对于每个气球,提供的输入是水平方向上,气球直径的开始和结束坐标.由于它是水平的,所以y坐标并不重要,因此只要知道开始和结束的x坐标就足够了.开始坐标总是小于结束坐标.平面内最多存在104个气球.一支弓箭可以沿着x轴从不同点完全垂直地射出.在坐标x处射出一支箭,若有一个气球的直径的开始和结束坐标为 xstart,xend, 且满足  xstart ≤ x ≤ xend,则该气球会被引爆.可以射出的弓箭的数量没有限制. 弓箭一旦被射出之后,可以无限地前进.我们想找到使得…
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…
463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…
Note: 后面数字n表明刷的第n + 1遍, 如果题目有**, 表明有待总结 Conclusion questions: [LeetCode] questions conclustion_BFS, DFS LeetCode questions conclustion_Path in Tree [LeetCode] questions conlusion_InOrder, PreOrder, PostOrder traversal [LeetCode] questions for Dynamic…
BFS, DFS 的题目总结. Directed graph: Directed Graph Loop detection and if not have, path to print all path. BFS/DFS: (可以用BFS或者DFS的,主要还是遍历) [LeetCode] 733. Flood Fill_Easy tag: BFS     1 [LeetCode] 690. Employee Importance_Easy tag: BFS    1 [LeetCode] 529…
突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…
DFS基础 深度优先搜索(Depth First Search)是一种搜索思路,相比广度优先搜索(BFS),DFS对每一个分枝路径深入到不能再深入为止,其应用于树/图的遍历.嵌套关系处理.回溯等,可以用递归.堆栈(stack)实现DFS过程. 关于广度优先搜索(BFS)详见:算法与数据结构基础 - 广度优先搜索(BFS) 关于递归(Recursion)详见:算法与数据结构基础 - 递归(Recursion) 树的遍历 DFS常用于二叉树的遍历,关于二叉树详见: 算法与数据结构基础 - 二叉查找树…