链表

基本技能

链表相关的核心点

null/nil 异常处理
dummy node 哑巴节点
快慢指针
插入一个节点到排序链表
从一个链表中移除一个节点
翻转链表
合并两个链表
找到链表的中间节点

常见题型

remove-duplicates-from-sorted-list

给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。

public class ListNode {
    public var val: Int
    public var next: ListNode?
    public init(_ val: Int) {
        self.val = val
        self.next = nil
    }
}

class Solution {
    //https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/solution/swift-bian-li-tong-shi-shan-chu-lian-biao-zhong-fu/
    func deleteDuplicates(_ head: ListNode?) -> ListNode? {
        var current = head
        while let now = current,let next = now.next{
            if now.val == next.val {
                now.next = next.next
            }else{
                current = now.next
            }
        }
        return head
    }
}

remove-duplicates-from-sorted-list-ii

给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中没有重复出现的数字。

思路：链表头结点可能被删除，所以用 dummy node 辅助删除

public class ListNode {
    public var val: Int
    public var next: ListNode?
    public init(_ val: Int) {
        self.val = val
        self.next = nil
    }
}

class Solution {
    //https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii/solution/swift-shuang-zhi-zhen-lian-biao-bian-li-tou-bu-tia/
    func deleteDuplicates(_ head: ListNode?) -> ListNode? {
        let top = ListNode(0)
        top.next = head
        var prev = top
        var current = head
        while let now = current, var next = now.next {
            if now.val == next.val{
                //求出next指向最后一个相等的
                while let end = next.next{
                    if now.val == end.val {
                        next = end
                    }else{
                        break
                    }
                }
                prev.next = next.next
                current = next.next
            }else{
                prev = now
                current = next
            }
        }
        return top.next
    }

    //递归实现
    //https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii/solution/swift-di-gui-shi-xian-by-hu-cheng-he-da-bai-sha/
    func deleteDuplicates(_ head: ListNode?) -> ListNode? {
        var res = head
        if let current = res,let next = current.next{
            if current.val == next.val{
                var end : ListNode? = next
                while end != nil  && current.val == end!.val {
                    end = end!.next
                }
                res = deleteDuplicates(end)
            }else{
                current.next = deleteDuplicates(next)
            }
        }
        return res
    }
}

注意点 • A->B->C 删除 B，A.next = C • 删除用一个 Dummy Node 节点辅助（允许头节点可变） • 访问 X.next 、X.value 一定要保证 X != nil

reverse-linked-list

反转一个单链表。

思路：用一个 prev 节点保存向前指针，temp 保存向后的临时指针

//https://leetcode-cn.com/problems/reverse-linked-list/solution/swift-shuang-zhi-zhen-bian-li-by-hu-cheng-he-da-ba/
//双指针遍历
//执行用时：20 ms, 在所有 Swift 提交中击败了78.48%的用户
//内存消耗：21.7 MB, 在所有 Swift 提交中击败了25.00%的用户
func reverseList_a(_ head: ListNode?) -> ListNode? {
    var head = head
    var prev: ListNode?
    while let current = head {
        let next = current.next
        current.next = prev
        prev = head
        head = next
    }
    return prev
}

//https://leetcode-cn.com/problems/reverse-linked-list/solution/swift-gao-xiao-di-gui-by-hu-cheng-he-da-bai-sha/
//递归
//执行用时：16 ms, 在所有 Swift 提交中击败了97.24%的用户
//内存消耗：22.1 MB, 在所有 Swift 提交中击败了25.00%的用户
func reverseList(_ head: ListNode?) -> ListNode? {
    guard let current = head,let next = current.next else {
        return head
    }
    let reverse = reverseList(next)
    next.next = current
    current.next = nil
    return reverse
}

reverse-linked-list-ii

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

思路：先遍历到 m 处，翻转，再拼接后续，注意指针处理

//https://leetcode-cn.com/problems/reverse-linked-list-ii/solution/swift-bian-li-fan-zhuan-by-hu-cheng-he-da-bai-sha/
//遍历反转
//执行用时：12 ms, 在所有 Swift 提交中击败了38.37%的用户
//内存消耗：20.9 MB, 在所有 Swift 提交中击败了100.00%的用户
func reverseBetween(_ head: ListNode?, _ m: Int, _ n: Int) -> ListNode? {
     let sentry = ListNode(0)
     sentry.next = head
     var prev: ListNode? = sentry
     var cur: ListNode? = head
     //移动到反转起始位置
     for _ in 1..<m{
         prev = cur
         cur = cur?.next
     }

     let beign = prev//记录反转第一个的前一个
     let end = cur//记录反转的第一个

     //反转m到n个元素
     for _ in m...n {
         let next = cur?.next
         cur?.next = prev
         prev = cur
         cur = next
     }

     beign?.next = prev//重新标记反转后的头
     end?.next = cur//重新标记反转后的尾
     return sentry.next
}

merge-two-sorted-lists

将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

思路：通过 dummy node 链表，连接各个元素

    //https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/swift-di-gui-by-hu-cheng-he-da-bai-sha/
    //递归实现
    //执行用时：16 ms, 在所有 Swift 提交中击败了94.75%的用户
    //内存消耗：21.1 MB, 在所有 Swift 提交中击败了100.00%的用户
    func mergeTwoLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        let res = ListNode()
        guard let cur1 = l1 else {
           return l2
        }
        guard let cur2 = l2 else {
           return l1
        }

        if cur1.val < cur2.val {
            res.next = cur1
            let next = mergeTwoLists(cur1.next,cur2)
            cur1.next = next
        }else{
            res.next = cur2
            let next = mergeTwoLists(cur1,cur2.next)
            cur2.next = next
        }

        return res.next
    }

    //https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/swift-die-dai-by-hu-cheng-he-da-bai-sha/
    //迭代
    //执行用时：16 ms, 在所有 Swift 提交中击败了94.75%的用户
    //内存消耗：20.8 MB, 在所有 Swift 提交中击败了100.00%的用户
    func mergeTwoLists_a(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var l1 = l1, l2 = l2
        let res = ListNode()
        var current = res
        while let cur1 = l1, let cur2 = l2 {
            if cur1.val < cur2.val{
                current.next = cur1
                l1 = cur1.next
            }else{
                current.next = cur2
                l2 = cur2.next
            }
            current = current.next!
        }

        current.next = l1 ?? l2

        return res.next
    }

partition-list

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

思路：将大于 x 的节点，放到另外一个链表，最后连接这两个链表

    //https://leetcode-cn.com/problems/partition-list/solution/swift-bian-li-headlian-biao-shi-shan-chu-xiao-yu-x/
    //遍历head链表时，删除小于x的所有元素，同时把他们生成一个小于x的链表，最后把小于x的链表加上遍历后的head链表
    //执行用时：16 ms, 在所有 Swift 提交中击败了71.00%的用户
    //内存消耗：20.8 MB, 在所有 Swift 提交中击败了100.00%的用户
    func partition(_ head: ListNode?, _ x: Int) -> ListNode? {
        let lessList = ListNode(0)//小于x的链表
        var lessTail = lessList //小于x的链表的尾部

        let res = ListNode(0)//原始链表的哨兵
        res.next = head
        var prev = res
        var cur = head

        while let item = cur {
            if item.val < x{
                prev.next = item.next
                lessTail.next = item
                lessTail = item
            }else{
                prev = item
            }
            cur = item.next
        }

        lessTail.next = res.next

        return lessList.next
    }

哑巴节点使用场景

当头节点不确定的时候，使用哑巴节点

sort-list

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。

思路：归并排序，找中点和合并操作

    //https://leetcode-cn.com/problems/sort-list/solution/swift-gui-bing-fei-di-gui-pai-xu-by-hu-cheng-he-da/
    //归并非递归排序
    //执行用时：284 ms, 在所有 Swift 提交中击败了84.55%的用户
    //内存消耗：24.7 MB, 在所有 Swift 提交中击败了100.00%的用户
    func sortList(_ head: ListNode?) -> ListNode? {
        //链表不存在或长度为1直接返回
        guard let item = head , item.next != nil else{
            return head
        }
        var count = 0 //链表长度
        var cur = head
        while cur != nil {
            count += 1
            cur = cur!.next
        }
        let dummy = ListNode(1)
        dummy.next = head

        //从长度为1个元素开始合并
        var length = 1
        while length < count {
            var begin = dummy
            var index = 0
            //遍历合并长度为length
            while index + length < count {
                var first = begin.next!,second: ListNode? = first
                var firstCount = length , secondCount = length
                //计算第二块的起始位置
                for _ in 0..<length{
                     second = second?.next
                 }

                //合并
                while firstCount > 0 && secondCount > 0 && second != nil {//注意第二块长度可能小于length
                    if first.val < second!.val {
                        begin.next = first
                        first = first.next!
                        firstCount -= 1
                    }else{
                        begin.next = second
                        second = second!.next
                        secondCount -= 1
                    }
                    begin = begin.next!
                }
                //第一块还有剩余
                while firstCount > 0 {
                     begin.next = first
                     first = first.next!
                     firstCount -= 1
                     begin = begin.next!
                }
                //第二块还有剩余
                while secondCount > 0 && second != nil {
                    begin.next = second
                    second = second!.next
                    secondCount -= 1
                    begin = begin.next!
                }
                begin.next = second//指向下次合并块的开始位置

                index += 2*length
            }
            length = 2*length
        }

        return dummy.next
    }

注意点

链表的递归函数调用空间复杂度：O(logn)，所以必须使用非递归

reorder-list

给定一个单链表 L：L→L→…→L__n→L 将其重新排列后变为： L→L__n→L→L__n→L→L__n→…

思路：找到中点断开，翻转后面部分，然后合并前后两个链表

    //https://leetcode-cn.com/problems/reorder-list/solution/swift-zhao-dao-zhong-dian-duan-kai-fan-zhuan-hou-m/
    //找到中点断开，翻转后面部分，然后合并前后两个链表
    //执行用时：96 ms, 在所有 Swift 提交中击败了82.22%的用户
    //内存消耗：26 MB, 在所有 Swift 提交中击败了100.00%的用户
    func reorderList(_ head: ListNode?) {
        guard head != nil && head!.next != nil else{
            return
        }

        //快慢指针寻找中间节点
        var slow = head!, fast = head!
        while fast.next != nil && fast.next!.next != nil  {
            slow = slow.next!
            fast = fast.next!.next!
        }
        var head2 = slow.next
        slow.next = nil

        //后半段反转
        var pre: ListNode? = nil
        while let cur = head2 {
            let next = cur.next
            cur.next = pre
            pre = cur
            head2 = next
        }
        head2 = pre

        //拼接
        let dummy = ListNode(0)

        var cur = dummy
        var h1 = head , h2 = head2
        while h2 != nil {
            cur.next  = h1

            let next1 = h1?.next
            h1?.next = h2
            h1 = next1

            cur = h2!
            h2 = h2?.next
        }

        cur.next = h1 //奇数时h1还有一个，偶数时是nil（是nil时也不影响）

        dummy.next = nil
    }

linked-list-cycle

给定一个链表，判断链表中是否有环。

思路：快慢指针，快慢指针相同则有环，证明：如果有环每走一步快慢指针距离会减 1

    //https://leetcode-cn.com/problems/linked-list-cycle/solution/swift-kuai-man-zhi-zhen-huan-lu-jian-ce-kong-jian-/
    //swift 快慢指针环路检测（ 空间复杂度O(1) ）
    //执行用时：64 ms, 在所有 Swift 提交中击败了92.48%的用户
    //内存消耗：22.2 MB, 在所有 Swift 提交中击败了100.00%的用户
    func hasCycle(_ head: ListNode?) -> Bool {
        guard let first = head , let second = first.next else{
            return false
        }

        var slow = first, fast: ListNode? = second
        while let item = fast {
            if slow === item {
                return true
            }
            slow = slow.next!
            fast = fast?.next?.next
        }
        return false
    }

linked-list-cycle-ii

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。

思路：快慢指针，快慢相遇之后，慢指针回到头，快慢指针步调一致一起移动，相遇点即为入环点

    //https://leetcode-cn.com/problems/linked-list-cycle-ii/solution/swift-kuai-man-zhi-zhen-kuai-man-xiang-yu-zhi-hou-/
    //swift 快慢指针，快慢相遇之后，慢指针回到头，快慢指针步调一致一起移动，相遇点即为入环点
    //执行用时：64 ms, 在所有 Swift 提交中击败了99.43%的用户
    //内存消耗：22 MB, 在所有 Swift 提交中击败了100.00%的用户
    func detectCycle(_ head: ListNode?) -> ListNode? {
        var slow = head, fast = head

        while fast != nil {
            slow = slow?.next
            fast = fast?.next?.next

            if slow === fast {
                slow = head
                while slow !== fast {
                    slow = slow?.next
                    fast = fast?.next
                }
                return slow
            }
        }

        return nil
    }

palindrome-linked-list

请判断一个链表是否为回文链表。

    //https://leetcode-cn.com/problems/palindrome-linked-list/solution/swift-kuai-man-zhi-zhen-zhao-dao-zhong-jian-jie--2/
    //快慢指针找到中间节点同时反转慢指针，之后比较前后两段（寻找中间节点判断更新）
    //执行用时：112 ms, 在所有 Swift 提交中击败了97.80%的用户
    //内存消耗：25.9 MB, 在所有 Swift 提交中击败了100.00%的用户
    func isPalindrome_a(_ head: ListNode?) -> Bool {
        if head == nil {
            return true
        }
        //快慢指针寻找中间节点,同时把前半部分反转
        var slow = head, fast = head
        var pre: ListNode? = nil
        while fast != nil && fast?.next != nil {
            let slowCur = slow
            slow = slow?.next
            fast = fast?.next?.next
            //下面是同时反转前部分
            slowCur?.next = pre
            pre = slowCur
        }


        var head1 = pre//第一段的头

        //第二段的头
        var head2: ListNode? = slow
        if fast != nil{//head总个是奇数时，调整第二部分头节点
            head2 = slow?.next
        }

        while let cur1 = head1, let cur2 = head2 {
            if cur1.val != cur2.val {
                return false
            }
            head1 = head1?.next
            head2 = head2?.next
        }
        return true
    }

copy-list-with-random-pointer

给定一个链表，每个节点包含一个额外增加的随机指针，该指针可以指向链表中的任何节点或空节点。要求返回这个链表的深拷贝。

思路：1、hash 表存储指针，2、递归， 3、复制节点跟在原节点后面

    //https://leetcode-cn.com/problems/copy-list-with-random-pointer/solution/swift-die-dai-tong-bu-shen-kao-bei-by-hu-cheng-he-/
    //迭代同步深拷贝
    //执行用时：48 ms, 在所有 Swift 提交中击败了90.16%的用户
    //内存消耗：22.3 MB, 在所有 Swift 提交中击败了100.00%的用户
    func copyRandomList(_ head: Node?) -> Node? {
        var visited = [UnsafeMutableRawPointer : Node]()

        func clonedNode(_ old: Node?) -> Node?{
            guard let from = old else{
                return nil
            }
            let key = Unmanaged.passUnretained(from).toOpaque()
            if let item = visited[key]{
                return item
            }
            let new = Node(from.val)
            visited[key] = new
            return new
        }

        var oldCur = head
        var newCur: Node?
        while let old = oldCur {
            let clonedItem = clonedNode(old)
            newCur?.next = clonedItem
            newCur = clonedItem
            newCur!.next = clonedNode(old.next)
            newCur!.random = clonedNode(old.random)
            oldCur = old.next
        }
        return clonedNode(head)
    }

    //https://leetcode-cn.com/problems/copy-list-with-random-pointer/solution/swift-shen-du-bian-li-di-gui-shi-xian-by-hu-cheng-/
    //深度遍历递归实现
    //执行用时：48 ms, 在所有 Swift 提交中击败了90.16%的用户
    //内存消耗：22.4 MB, 在所有 Swift 提交中击败了100.00%的用户
    var visited = [UnsafeMutableRawPointer : Node]()
    func copyRandomList_a(_ head: Node?) -> Node? {
        guard let from = head else {
            return nil
        }

        let key = Unmanaged.passUnretained(from).toOpaque()
        if let item = visited[key]{
                return item
        }

        let copedItem = Node(from.val)
        visited[key] = copedItem
        copedItem.random = copyRandomList_a(from.random)
        copedItem.next = copyRandomList_a(from.next)

        return copedItem
    }

    //https://leetcode-cn.com/problems/copy-list-with-random-pointer/solution/swift-fu-zhi-jie-dian-gen-zai-yuan-jie-dian-hou-mi/
    //swift 复制节点跟在原节点后面（空间O(1)）
    //执行用时：48 ms, 在所有 Swift 提交中击败了90.16%的用户
    //内存消耗：21.7 MB, 在所有 Swift 提交中击败了100.00%的用户
    func copyRandomList_b(_ head: Node?) -> Node? {
           guard head != nil else {
               return nil
           }

           var cur = head
           while let old = cur {
            let copied = Node(old.val)
            copied.next = old.next
            old.next = copied
            cur = copied.next
           }

           cur = head
           while let old = cur {
            old.next?.random = old.random?.next
            cur = old.next?.next
           }

           cur = head
           let copiedList = head!.next
           var pre: Node? = nil
           while let old = cur {
                     pre?.next = old.next
                     pre = old.next

                     old.next = old.next?.next

                     cur = old.next
            }

            return copiedList
       }

总结

链表必须要掌握的一些点，通过下面练习题，基本大部分的链表类的题目都是手到擒来~

null/nil 异常处理
dummy node 哑巴节点
快慢指针
插入一个节点到排序链表
从一个链表中移除一个节点
翻转链表
合并两个链表
找到链表的中间节点

练习

最后更新于5年前

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