本文将使用Swift实现一个标准链表,在实现的过程中,遵守函数式编程的规则,无副作用,可以看到和C语言的实现还是有较大的差异. 预备知识 enum 的各种用法 swift的基本的模式匹配( pattern matching ) -- 代码里面 case 后面部分属于模式匹配,包含 switch case , guard case , let case 等 extension 的用法 guard generic 遇到不懂的地方可以参阅官方文档学习即可 链表定义 首先来看一下Wikipedia对Li

最近在回答StackOverflow的问题时,发现performSelector方法在Swift被去掉,Apple的注释是这个方法被去掉是因为不安全: NOTE The performSelector: method and related selector-invoking methods are not imported in Swift because they are inherently unsafe. 如果在Swift调用这个方法会编译出错: 'performSelector' is

在OC里使用惯了AFNetworking,比较喜欢这一个第三方库,在别的途径里得知可以在Swift里使用AFNetworking.但是那个时候我不知道具体的操作是怎样的,于是我只能去百度.GOOGLE了,还好让我给找到了答案,因而在这里和大家分享一下. 1.首先把下载好的AFN库直接拖到工程中 2.这里把Copy item if needed这个选项勾上,然后点完成 3.这时会有一个弹出框,点击YES 4.这个是拖拽好的截图,你会发现多了一个Header文件,也就是图中高亮的那个文件了.因为我的

Swift里对于字符串这些引入了index型,相对其他语言而言字符操作更安全了,但是问题就是一不注意搞错范围就会有各种离奇的bug. 在讲主题前,先说个小细节. Swift里非常严密的定义了一大堆字符串长度量,但我们实际开发基本上只有字符(character)的长度有用. 还是上例子 var strTemp = "这只是一个萌萌哒小测试" print("萌萌哒长度: " + String(strTemp.characters.count)) 运行结果如下: 萌萌哒长

中缀语法是OC里特有的一种,就是在函数的参数前面加一个解释词,让调用的时候明白该参数的含义 比如: -(void)processDataWithparamaA:(NSString *)paramaA paramaB:(NSString *)paramaB{ } 调用的时候: [self processDataWithparamaA:@"A" paramaB:@"B"]; 不过你发现没,这中缀语法的前提是你必须在函数名的最前面,写个With啥的说明第一个参数的名字,否

下午在适配iPadUI的时候,用到了UIPopoverPresentationController,然后在转屏的时候需要调用UIPopoverPresentationControllerDelegate来返回一个适配后的view和CGRect,这里先看下在OC里的写法: - (void)popoverPresentationController: (nonnull UIPopoverPresentationController *) popoverPresentationController w

init() 函数 在 Array 里 public init() { _buffer = _Buffer() } 以Buffer 是 _ContiguousArrayBuffer 为例. 即初始化了一个_ContiguousArrayBuffer. 在_ContiguousArrayBuffer里 /// Create an empty buffer. @inlinable internal init() { _storage = _emptyArrayStorage } 其中 _emptyA

以append操作为例 public mutating func append(_ other: String) { if self.isEmpty && !_guts.hasNativeStorage { self = other return } self._guts.append(other._guts) } _StringGuts 做了实际的工作 下面是实际进行append的地方 internal mutating func append(_ slicedOther: _Strin

Given a linked list, remove the n-th node from the end of list and return its head. Example: Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Given n w

Given a linked list, rotate the list to the right by k places, where k is non-negative. Example 1: Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3-&g

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. Example: Input: head = 1->4->3-

Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list. Example: Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL 反转从位置 m 到 n 的链表.请使用一趟扫描完成反转. 说明:1 ≤ m ≤ n ≤ 链表长度.

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never diffe

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? 给定一个链表,返回链表开始入环的第一个节点. 如果链表无环,则返回 null. 说明:不允许修改给定的链表. 进阶:你是否可以

Given a singly linked list L: L0→L1→…→Ln-1→Ln,reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… You may not modify the values in the list's nodes, only nodes itself may be changed. Example 1: Given 1->2->3->4, reorder it to 1->4->2->3. Example 2: G

Sort a linked list using insertion sort. A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list.With each iteration one element (red) is removed from the input data and inserted in

Sort a linked list in O(n log n) time using constant space complexity. Example 1: Input: 4->2->1->3 Output: 1->2->3->4 Example 2: Input: -1->5->3->4->0 Output: -1->0->3->4->5 在 O(n log n) 时间复杂度和常数级空间复杂度下,对链表进行排序.

Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3 begin to intersect at node c1. Notes: If the two linked lists have

Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? 反转一个单链表. 示例: 输入: 1->2->

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. You should try to do it in place. The program should run in O(1) space complexi