链表问题集锦（二）

悠悠 2022-05-31 01:21 167阅读 0赞

原文站点：[https://senitco.github.io/2018/02/02/data-structure-linkedlist-2/][https_senitco.github.io_2018_02_02_data-structure-linkedlist-2]

数据结构中经典链表问题总结归纳，接上一篇[链表问题集锦（一）][Link 1]

### 单链表逆转 ###

题目描述：给定一个单链表，将其逆转。[LeetCode][]  
分析：迭代法求解，借助三个指针start、head、next，start指向逆转后（已重排）的第一个结点，head指向未重排的第一个结点，next指向head的下一个结点，每次将未重排的第一个结点插入到已重排序列的首部。也可利用递归法实现。

ListNode* reverseList(ListNode* head) 
    {
        ListNode *start = NULL, next;
        while(head)
        {
            next = head->next;
            head->next = start;
            start = head;
            head = next;
        }
        return start;
    }
    
    //增加一个头结点的迭代版本
    ListNode* reverseList(ListNode* head) 
    {
        if(!head)
            return NULL;
        ListNode* start = new ListNode(0);
        start->next = head;
        ListNode *first = head;
        ListNode *second = first->next;
        while(second)
        {
            first->next = second->next;
            second->next = start->next;
            start->next = second;
            second = first->next;
        }
        return start->next;
    }
    
    //递归法
    ListNode* reverseList(ListNode* head) 
    {
        return recursive(head, NULL);
    }
    
    ListNode* recursive(ListNode* head, ListNode* start)
    {
        if(!head)   return start;
        ListNode* next = head->next;
        head->next = start;
        return recursive(next, head);
    }
    
    //直接递归
    Node* reverseByRecursion(Node *head)
    {
        //第一个条件是判断异常，第二个条件是结束判断
        if(head == NULL || head->next == NULL) 
            return head;
        Node *newHead = reverseByRecursion(head->next);
        head->next->next = head;
        head->next = NULL;
        return newHead;    //返回新链表的头指针
    }

### 将单链表进行分组逆转 ###

问题描述：将单链表按顺序分成N组，每组K个元素，最后一组可能少于K个，分别将每组元素逆转，并链接成一个完整的序列。例如一个链表：1->2->3->4->5，如果k=2，则逆转后的序列为2->1->4->3->5；如果k=3，则逆转序列为3->2->1->4->5。[LeetCode][LeetCode 1]  
分析：分别对每组结点进行逆转，可用迭代法和递归法

/**迭代法：
    -1 -> 1 -> 2 -> 3 -> 4 -> 5
     |    |    |     
    pre  cur  nex  
    
    -1 -> 2 -> 1 -> 3 -> 4 -> 5
     |         |    |     
    pre       cur  nex  
    
    -1 -> 3 -> 2 -> 1 -> 4 -> 5
     |              |    |    
    pre            cur  nex 
    
    三个重要指针，头指针pre，始终指向已重新排列的第一个元素；当前指针cur，指向已排序的最后一个元素；
    cur->next则指向未排序的第一个元素也就是nex指针，nex的作用是即将作为第一个元素赋给pre->next.
    */
    ListNode* reverseKGroup(ListNode* head, int k) 
    {
        if(head == nullptr || k == 1)
            return head;
        int num = 0;
        ListNode* preheader = new ListNode(0);
        ListNode *pre, *nex, *cur;
        preheader->next = head;
        cur = pre = preheader;
        while(cur = cur->next)
            num++;
        while(num >= k)
        {
            cur = pre->next;
            nex = cur->next;
            for(int i = 1; i < k; i++)
            {
                cur->next = nex->next;
                nex->next = pre->next;
                pre->next = nex;
                nex = cur->next;
            }
            pre = cur;
            num -= k;
        }
        return preheader->next;
    }
    
    /**递归法：
    head：指向正序起始节点
    curr：指向反转后的起始节点
    tmp：head->next，用于保存下一个插入反转序列的节点
    */
    ListNode* reverseKGroup(ListNode* head, int k) 
    {
        ListNode* curr = head;
        int count = 0;
        while (curr != nullptr && count != k) { // find the k+1 node
            curr = curr->next;
            count++;
        }
        if (count == k) { // if k+1 node is found
            curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
            // head - head-pointer to direct part, 
            // curr - head-pointer to reversed part;
            while (count-- > 0) { // reverse current k-group: 
                ListNode* tmp = head->next; // tmp - next head in direct part
                head->next = curr; // preappending "direct" head to the reversed list 
                curr = head; // move head of reversed part to a new node
                head = tmp; // move "direct" head to the next node in direct part
            }
            head = curr;
        }
        return head;
    }

### 将单链表的结点两两交换 ###

题目描述：给定一个单链表，将相邻两个结点两两交换。例如链表：1->2->3->4, 交换后的序列为2->1->4->3。[LeetCode][LeetCode 2]

//遍历元素，两两交换
    ListNode* swapPairs(ListNode* head) 
    {
        ListNode* start = new ListNode(0);
        start->next = head;
        ListNode* current = start;
        ListNode *first, *second;
        while(current->next && current->next->next)
        {
            first = current->next;
            second = first->next;
            first->next = second->next;
            current->next = second;
            second->next = first;
            current = current->next->next;
        }
        return start->next;
    }
    
    //递归法
    ListNode* swapPairs(ListNode* head) 
    {
        if(head == nullptr || head->next == nullptr)
            return head;
        ListNode* pNode = head->next;
        head->next = swapPairs(head->next->next);
        pNode->next = head;
        return pNode;
    }
    
    //二级指针
    ListNode* swapPairs(ListNode* head) 
    {
        ListNode **pp = &head, *a, *b;
        while ((a = *pp) && (b = a->next)) 
        {
            a->next = b->next;
            b->next = a;
            *pp = b;    //改变指针指指向的地址
            pp = &(a->next);
        }
        return head;
    }

### 按特定序列将单链表重排 ###

问题描述：Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…  
You must do this in-place without altering the nodes’ values.  
For example, Given \{1,2,3,4\}, reorder it to \{1,4,2,3\}.[LeetCode][LeetCode 3]  
分析：分三步走：  
(1)找到链表的中点位置middle=(len-1)/2  
(2)将后半部分（middle之后的元素）链表逆转  
(3)同时遍历两部分链表元素，依次将后半部分元素插入到前半部分的相应位置

void reorderList(ListNode* head) 
    {
        if(!head || !head->next)
            return;
        ListNode *p1, *p2;
        p1 = p2 = head;
        //Find the middle of the list, 1->2->3->4->5->6, middle point to 3
        while(p2->next && p2->next->next)
        {
            p1 = p1->next;
            p2 = p2->next->next;
        }
        //Reverse the half after middle  1->2->3->4->5->6 to 1->2->3->6->5->4
        ListNode *middle = p1, *prev = p1->next, *cur;
        while(prev->next)
        {
            cur = prev->next;
            prev->next = cur->next;
            cur->next = middle->next;
            middle->next = cur;
        }
        //Start reorder one by one  1->2->3->6->5->4 to 1->6->2->5->3->4
        p1 = head;
        p2 = middle->next;
        while(p1 != middle)
        {
            middle->next = p2->next;
            p2->next = p1->next;
            p1->next = p2;
            p1 = p2->next;
            p2 = middle->next;
        }
    }

### 合并两个有序链表 ###

问题描述：将两个有序链表合并成一个新的有序链表。[LeetCode][LeetCode 4]

//迭代法
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) 
    {
        ListNode* start = new ListNode(0);
        ListNode *pNode = start;
        while(l1 && l2)
        {
            if(l1->val > l2->val)
            {
                pNode->next = l2;
                l2 = l2->next;
            }
            else
            {
                pNode->next = l1;
                l1 = l1->next;
            }
            pNode = pNode->next;
        }
        if(l1)  
            pNode->next = l1;
        else
            pNode->next = l2;
    
        return start->next;
    }
    
    //递归法
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) 
    {
        if(l1 == NULL) return l2;
        if(l2 == NULL) return l1;
    
        if(l1->val < l2->val) 
        {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } 
        else 
        {
            l2->next = mergeTwoLists(l2->next, l1);
            return l2;
        }
    }

### 合并K个有序链表 ###

问题描述：[LeetCode][LeetCode 5]

/**
    法一：与合并两个有序列表的思想一致，每次比较 K 个节点，取其最小值；在遍历过程中，
    若有列表已到达末尾节点，则将该列表从vector中移除
    */
    ListNode* mergeKLists(vector<ListNode*>& lists) 
    {
        if (lists.empty())
            return NULL;
        vector<ListNode*> pl;
        for (auto node : lists)
            pl.push_back(node);
        ListNode* start = new ListNode(0);
        ListNode* pNode = start;
        int value = INT_MAX, index = 0;
        while (true)
        {
            for (auto iter = pl.begin(); iter != pl.end();)
            {
                if (!(*iter))
                {
                    iter = pl.erase(iter);  //注意erase()可能会造成后面的迭代器失效
                }
                else
                    iter++;
            }
            if (pl.empty())
                break;
            value = INT_MAX;
            index = 0;
            for (int i = 0; i < pl.size(); i++)
            {
                if (pl[i]->val < value)
                {
                    value = pl[i]->val;
                    index = i;
                }
            }
            pNode->next = pl[index];
            pl[index] = pl[index]->next;
            pNode = pNode->next;
        }
        return start->next;
    }
    
    //法二：借助优先队列实现
    struct cmp
    {
        bool operator()(ListNode* l1, ListNode* l2)
        {
            return l1->val > l2->val;
        }
    };
    
    ListNode* mergeKLists(vector<ListNode*>& lists) 
    {
        if (lists.empty())
            return NULL;     
        ListNode* start = new ListNode(0);
        ListNode* pNode = start;
        priority_queue<ListNode*, vector<ListNode*>, cmp> pql;
        for (auto node : lists)
            if(node)            //注意此处要判断节点(链表)是否为空
                pql.push(node);
        while(!pql.empty())
        {
            pNode->next = pql.top();
            pql.pop();
            pNode = pNode->next;
            if(pNode->next != NULL)
                pql.push(pNode->next);
        }     
        return start->next;
    }
    
    //法三：递归求解，将有序列表两两合并
    ListNode *mergeKLists(vector<ListNode *> &lists) 
    {
        if(lists.empty()){
            return nullptr;
        }
        while(lists.size() > 1){    
            lists.push_back(mergeTwoLists(lists[0], lists[1])); //mergeTwoLists()见上
            lists.erase(lists.begin());
            lists.erase(lists.begin());
        }
        return lists.front();
    }

[https_senitco.github.io_2018_02_02_data-structure-linkedlist-2]: https://senitco.github.io/2018/02/02/data-structure-linkedlist-2/
[Link 1]: https://senitco.github.io/2018/02/01/data-structure-linkedlist/
[LeetCode]: https://leetcode.com/problems/reverse-linked-list/description/
[LeetCode 1]: https://leetcode.com/problems/reverse-nodes-in-k-group/description/
[LeetCode 2]: https://leetcode.com/problems/swap-nodes-in-pairs/description/
[LeetCode 3]: https://leetcode.com/problems/reorder-list/description/
[LeetCode 4]: https://leetcode.com/problems/merge-two-sorted-lists/description/
[LeetCode 5]: https://leetcode.com/problems/merge-k-sorted-lists/description/