[LeetCode] 24. Swap Nodes in Pairs 两两交换链表中的节点

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)

Example 1:

Input: head = [1,2,3,4]
Output: [2,1,4,3]

Example 2:

Input: head = []
Output: []

Example 3:

Input: head = [1]
Output: [1]

Constraints:

The number of nodes in the list is in the range [0, 100].
0 <= Node.val <= 100

这道题不算难，是基本的链表操作题，我们可以分别用递归和迭代来实现。对于迭代实现，还是需要建立 dummy 节点，注意在连接节点的时候，最好画个图，以免把自己搞晕了，参见代码如下：

解法一：

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

递归的写法就更简洁了，实际上利用了回溯的思想，递归遍历到链表末尾，然后先交换末尾两个，然后依次往前交换：

解法二：

class Solution {
public:
    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;
    }
};