pointers专题

Leetcode211: Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant e

PAT甲级 1085 Perfect Sequence 二分和双指针(Two Pointers)

二分写法 #include <bits/stdc++.h>using namespace std;int find_upper_bound(const vector<long long>& nums, long long x){int beg = 0, end = nums.size(), mid = beg + (end - beg) / 2;while (beg < end) {mid

LeetCode - 双指针(Two Pointers) 算法集合 [对撞指针、快慢指针、滑动窗口、双链遍历]

欢迎关注我的CSDN:https://spike.blog.csdn.net/ 本文地址:https://spike.blog.csdn.net/article/details/139270999 双指针算法是一种常见且灵活的技巧,通过使用两个指针协同完成任务。这些指针可以指向不同的元素,具体应用取决于问题的性质。双指针算法的常见用法: 对撞指针:一左一右向中间逼近。例如,反转字符串

populating-next-right-pointers-in-each-node-ii

1、populating-next-right-pointers-in-each-node-ii 来源:牛客网 Follow up for problem "Populating Next Right Pointers in Each Node".What if the given tree could be any binary tree? Would your previous solut

LeetCode - populating-next-right-pointers-in-each-node

题目: Given a binary tree struct TreeLinkNode {TreeLinkNode *left;TreeLinkNode *right;TreeLinkNode *next;}   Populate each next pointer to point to its next right node. If there is no next righ

理解Pointers In C++:第一重

Variable vs. Pointer int foo;int *foo_ptr = &foo; Grammatrically speaking, there is no such thing as a “pointer variable”: all variables are the same. There are, however, variables with differe

序列合并问题---two pointers

题目描述: 假设有两个递增序列A和B,要 求将它们合并为一个递增序列C 核心代码实现 int merge(int A[],int B[],int C[],int n,int m){int i=0,j=0,index=0;//i指向A[0],j指向B[0]while(i<n&&j<m){if(A[i]<=B[j]){C[index++]=A[i];//将A[i]加入序列C}else

LeetCode42题,单调栈、构造法、two pointers,这道Hard题的解法这么多?

本文始发于个人公众号:TechFlow,原创不易,求个关注 今天是LeetCode专题的第23篇文章。 今天来看一道很有意思的题,它的难度是Hard,并且有许多种解法。 首先我们来看题面,说是我们有若干个水坝,水坝的宽都是1,但是水坝的高度参差不齐。某一天我们向水坝围起来的部分灌水,一直到灌满为止,请问水坝中存储了多少单位的水?我们可以参考一下下图: 上图当中黑色的部分是水坝,蓝色

LeetCode 117 Populating Next Right Pointers in Each Node II (链表 层次遍历 推荐)

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra

C++高级面试题:什么是 C++ 中的多态指针(Polymorphic Pointers)?

什么是 C++ 中的多态指针(Polymorphic Pointers)? 在 C++ 中,多态指针(Polymorphic Pointers)通常指向基类(Base Class)的指针,但它可以指向派生类(Derived Class)的对象。多态指针允许在运行时根据对象的实际类型来调用相应的函数,从而实现多态性(Polymorphism)。 多态指针是通过将基类指针指向派生类对象来实现的。这

leetcode 117 Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant ext

leetcode117~Populating Next Right Pointers in Each Node II

Follow up for problem “Populating Next Right Pointers in Each Node”. What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra

常量和指针(Pointers and Constants)

常量和指针( Pointers and Constants ) ——const 修饰的指针解惑 一般遇到用const修饰的常量涉及到指针就会比较麻烦,容易把头搞晕,有个简单的技巧就是从右向左看,下面我举例子说明: const int* p1 = NULL; //写法一 int const* p2 = NULL; //写法二 int *const p

leetcode----117. Populating Next Right Pointers in Each Node

链接: https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ 大意: 大意同 完全二叉树版。 只不过在本题中树并不是完全二叉树 思路: 思路与完全二叉树版一模一样。 代码: 代码与完全二叉树版一模一样。 结果: 结论: 层次遍历递归实现。

leetcode----116. Populating Next Right Pointers in Each Node

链接: https://leetcode.com/problems/populating-next-right-pointers-in-each-node/ 大意: 给定一棵完全二叉树root,该二叉树的所有叶子节点都在同一层。该二叉树数据结构定义如下: class Node {public int val;public Node left;public Node right;public

fast-slow-pointers

快慢指针     快慢指针中的快慢指的是移动的步长,即每次向前移动速度的快慢。例如可以让快指针每次沿链表向前移动2,慢指针每次向前移动1次。 快慢指针的应用 (1)判断单链表是否存在环 如果链表存在环,就好像操场的跑道是一个环形一样。此时让快慢指针都从链表头开始遍历,快指针每次向前移动两个位置,慢指针每次向前移动一个位置;如果快指针到达NULL,说明链表以NULL为结尾,没有环。如果快指针追

leetcode-Populating Next Right Pointers in Each Node

题目描述 Given a binary tree struct TreeLinkNode {TreeLinkNode *left;TreeLinkNode *right;TreeLinkNode *next;} Populate each next pointer to point to its next right node. If there is no next

LeetCode之Populating Next Right Pointers in Each Node II

/*由于二叉树的层序遍历空间是O(n),可以利用建立的链表进行遍历。参考自:https://github.com/soulmachine/leetcode*/class Solution {public:void connect(TreeLinkNode *root) {while(root != nullptr){TreeLinkNode *pre(nullptr), *next(nullp

Smart pointers

Smart pointers基本模型 template <class T> class SmartPtr { public: explicit SmartPtr(T* pointee):pointee_(pointee); SmartPtr &operator=(const SmartPtr&other); ~SmartPtr(); T& operator

关于error C2110: cannot add two pointers的处理

当出现error C2110: cannot add two pointers时,是因为字符串相加不能以常量开头,举例如下:strCheckSum = FREAM_LENGHT+ FREAM_TYPE + strResult; FREAM_LENGHT、FREAM_TYPE 都是字符串常量宏,如果按照上面方式书写代码会出现上面的错误。改为:strCheckSum = strCheckSum + F

LeetCode 之双指针 two pointers

1. 3Sum Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: Elements in a triplet

[LeetCode73]Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant e

【more effective c++读书笔记】【第5章】技术(4)——Smart Pointers(智能指针)

一、类似C++标准程序库提供的auto_ptr template的智能指针 //Smart.h#ifndef SMART_H#define SMART_Htemplate<typename T>class SmartPtr{public:SmartPtr(T* realPtr = 0); //构造函数~SmartPtr();//析构函数SmartPtr(SmartPtr& rhs);

【C++ techniques】Smart Pointers智能指针

Smart Pointers智能指针 看起来、用起来、感觉起来像内置指针,但提供更多的机能。拥有以下各种指针行为的控制权: 构造和析构;复制和赋值;解引。 Smart Pointers的构造、赋值、析构 C++的标准程序库提供的auto_ptr template: auto_ptr对象是个smart pointer,用来指向诞生于堆内的对象,直到该auto_ptr被销毁为止;当销毁发生时,