Swap Nodes in Pairs leetcode

最新推荐文章于 2025-09-10 15:24:29 发布

转载最新推荐文章于 2025-09-10 15:24:29 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/sdlwlxf/p/5122770.html

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#数据结构与算法

本文介绍了一种链表操作的高效方法，通过使用双重指针优化相邻节点交换过程，节省了额外空间并提高了算法效率。

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.

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未优化的代码，用3个临时指针

ListNode* swapPairs(ListNode* head) {
    if (head == nullptr || head->next == nullptr)
        return head;
    ListNode *pre = head;
    ListNode *cur = pre->next;
    ListNode *net = cur->next;
    head = cur;
    while (true)
    {
        cur->next = pre;
        if (net == nullptr || net->next == nullptr) {
            pre->next = net;
            break;
        }        
        pre->next = net->next;
        pre = net;
        cur = pre->next;        
        net = cur->next;    
    }
    return head;
}

惊人的解法，用双重指针，开阔思路，本以为自己可以随心所欲使用双指针，看来还是差得远

ListNode *swapPairs(ListNode *head) {
    ListNode **p = &head;

    while (*p && (*p)->next) {
        ListNode *t = (*p)->next;

        (*p)->next = t->next;
        t->next = *p;
        *p = t;

        p = &(*p)->next->next;
    }

    return head;
}