Cinte's Leetcode Record
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25. K 个一组翻转链表

Posted on Edited on In Disqus:
Symbols count in article: 2.6k Reading time ≈ 2 mins.

25. K 个一组翻转链表

递归

递归栈需要O(n / k)的空间复杂度

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(!head)
            return nullptr;
        auto tail = head;
        ListNode* pre = nullptr;
        ListNode* pt = head;
        for(int i = 0; i < k; ++i)
            if(!pt)
                return head;
            else
                pt = pt->next;
        for(int i = 0; i < k; ++i)
        {
            if(!head)
                break;
            auto next = head->next;
            head->next = pre;
            pre = head;
            head = next;
        }
        tail->next = reverseKGroup(head, k);
        return pre;
    }
};

迭代

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* pseudoHead = new ListNode();
        pseudoHead->next = head;
        ListNode* tail;
        auto pre = pseudoHead;
        while(pre)
        {
            ListNode* hair = pre;
            for(int i = 0; i < k; ++i)
            {
                pre = pre->next;
                if(!pre)
                    return pseudoHead->next;
            }
            auto next = pre->next;
            tie(head, tail) = reverse(hair->next, next);
            hair->next = head;
            tail->next = next;
            pre = tail;
        }
        return pseudoHead->next;
    }
private:
    pair<ListNode*, ListNode*> reverse(ListNode* head, ListNode* tail)
    {
        ListNode* pre = nullptr;
        ListNode* pt = head;
        while(pt != tail)
        {
            auto next = pt->next;
            pt->next = pre;
            pre = pt;
            pt = next;
        }
        return {pre, head};
    }
};

穿针引线

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(k == 1)
            return head;
        ListNode ret;
        ListNode *dummyHead{&ret};
        ret.next = head;
        auto pre = dummyHead;
        int count{k};
        while(head)
        {
            auto tmp = head;
            int c = k;
            while(c--)
            {
                if(!tmp)
                    return ret.next;
                tmp = tmp->next;
            }
            while(--count)
            {
                auto next = head->next;
                if(next)
                {
                    head->next = next->next;
                    next->next = pre->next;
                }
                else
                    head->next = nullptr;
                pre->next = next;
            }
            pre = head;
            head = head->next;
            count = k;
        }
        return ret.next;
    }
};