Swap Nodes in Pairs

最新推荐文章于 2019-05-11 21:02:45 发布

转载最新推荐文章于 2019-05-11 21:02:45 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/Kai-Xing/p/4101429.html

本文介绍了一种链表中相邻节点的交换算法，通过递归和迭代两种方式实现。该算法仅使用常数空间，不修改列表中的值，只改变节点指向。递归方法简洁明了，而迭代方法更易于理解。

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.

分析：很容易想到的方式是递归，时间复杂度递推公式为T(n) = T(n-2) + 2，可知时间复杂度为O(n)。代码：

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        if(head == NULL || head->next == NULL) return head;
        
        ListNode * tail = swapPairs(head->next->next);
        ListNode * tmp = head->next;
        head->next = tail;
        tmp->next = head;
        
        return tmp;
    }
};

迭代版：

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        if(head == NULL || head->next == NULL) return head;
        
        ListNode * dummy = new ListNode(-1);
        dummy->next = head;
        
        for(ListNode * pre_p = dummy, *p = head; p && p->next;){
            ListNode * tmp = p->next;
            p->next = p->next->next;
            tmp->next = p;
            pre_p->next = tmp;
            
            pre_p = p;
            p = p->next;
        }
        
        return dummy->next;
    }
};