[ LeetCode ] #24. Swap Nodes in Pairs（交换相邻节点 C++ & Python）

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.

题意：

给一个单链表，返回相邻节点相互交换后的新链表。

分析：

如果不注意上面大大的note ，直接依次遍历交换相邻节点的值，即可得到结果。如Code1.

但既然题目要求必须使用交换节点，因此就需要对Code1进行相应的更改。

方法2：

以下为A，B节点交换的过程。pre用来保存当前交换节点的前驱节点，tmp指向当前交换前的左元素，post保存交换前tmp的下一元素。

使用临时变量post作为tmp的下一节点。
使tmp 的前一节点的 next 指向post。
使tmp的next指向post的下一节点，即可与后面未处理的链表连接起来。
使pre指向tmp，tmp指向tmp->next，即更新交换的位置。交换过程如下图。实现见Code2.

[ LeetCode ] #24. Swap Nodes in Pairs（交换相邻节点 C++ & Python）

Code1 & C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if(head==NULL)return head;
        ListNode *tmp = head;
        while(tmp->next != NULL){
            ListNode *t = tmp->next;
            swap(tmp->val, t->val);
            if(t->next!=NULL)
                tmp = t->next;
            else
                break;
        }
        return head;
    }
};

结果：

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Swap Nodes in Pairs.

Memory Usage: 9.1 MB, less than 5.19% of C++ online submissions for Swap Nodes in Pairs.

Code2 & C++
在实行交换之前，说先对前两个元素进行了处理，因为这里头元素没有前驱即为第一个元素，没有前驱节点，因此，这里简单地作为单独处理。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if(head==NULL || head->next==NULL)return head;
        ListNode *pre = head;
        ListNode *tmp = pre->next;
        pre->next = tmp->next;
        tmp->next = pre;
        head = tmp;
        tmp = pre->next;

        while(tmp!=NULL && tmp->next!=NULL){
            ListNode *post = tmp->next;
            pre->next = post;
            tmp->next = post->next;
            post->next = tmp;
            pre = tmp;
            tmp = pre->next;
        }
        return head;
    }
};

结果：

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Swap Nodes in Pairs.

Memory Usage: 8.9 MB, less than 44.23% of C++ online submissions for Swap Nodes in Pairs.

Code2 & Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:return head
        tmp = head.next
        head.next = tmp.next
        tmp.next = head
        head = tmp
        pre = tmp.next
        tmp = pre.next
        while tmp is not None and tmp.next is not  None:
            post = tmp.next
            pre.next = post
            tmp.next = post.next
            post.next = tmp
            pre = tmp
            tmp = tmp.next
        return head

结果：

Runtime: 36 ms, faster than 74.40% of Python3 online submissions for Swap Nodes in Pairs.

Memory Usage: 13.2 MB, less than 5.04% of Python3 online submissions for Swap Nodes in Pairs.

[ LeetCode ] #24. Swap Nodes in Pairs（交换相邻节点 C++ & Python）

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