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24. Swap Nodes in Pairs

24. Swap Nodes in Pairs

作者: weego | 来源:发表于2018-04-07 09:54 被阅读0次

    Description

    Given a linked list, swap every two adjacent nodes and return its head.

    For example,
    Given 1->2->3->4, you should return the list as 2->1->4->3.
    

    Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

    Solution

    medium有点扯,注意链表指针的断裂次序

    ListNode* swapPairs(ListNode* head) {
        auto dummy = new ListNode(-1), pNode = dummy;
        dummy->next = head;
        while (pNode->next && pNode->next->next) {
            auto pNext = pNode->next;
            pNode->next = pNext->next;
            pNext->next = pNext->next->next;
            pNode->next->next = pNext;
            pNode = pNext;
        }
        return dummy->next;
    }
    

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