[LintCode] Swap Nodes in Pairs 成对交换节点

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

Example

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

Challenge

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

LeetCode上的原题，请参见我之前的博客Swap Nodes in Pairs。

解法一：

class Solution {
public:
    /**
     * @param head a ListNode
     * @return a ListNode
     */
    ListNode* swapPairs(ListNode* head) {
        ListNode *dummy = new ListNode(-1), *pre = dummy;
        dummy->next = head;
        while (pre->next && pre->next->next) {
            ListNode *t = pre->next;
            pre->next = t->next;
            t->next = t->next->next;
            pre->next->next = t;
            pre = t;
        }
        return dummy->next;
    }
};

解法二：

class Solution {
public:
    /**
     * @param head a ListNode
     * @return a ListNode
     */
    ListNode* swapPairs(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode *t = head->next;
        head->next = swapPairs(head->next->next);
        t->next = head;
        return t;
    }
};