[LeetCode]24.SwapNodesinPairs成对交换节点

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

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example:

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

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，***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;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/24

类似题目：

Reverse Nodes in k-Group

参考资料：

https://leetcode.com/problems/swap-nodes-in-pairs

https://leetcode.com/problems/swap-nodes-in-pairs/discuss/11030/My-accepted-java-code.-used-recursion.

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