【leetcode】24. Swap Nodes in Pairs 链表奇偶节点交换

1. 题目

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.

2. 思路

遍历每次交换两个。
注意：两个相邻pair之间的链要注意更新好，比如1->4.
其实更简单的是直接递归完成，这样只需要处理前两个节点。

3. 代码

耗时：3ms

/**
 * 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* p1 = head;
        ListNode* p2 = p1->next;
        while (p2 != NULL) {
            int tmp = p1->val;
            p1->val = p2->val;
            p2->val = tmp;
            p1 = p2->next;
            p2 = NULL;
            if (p1 != NULL) {
                p2 = p1->next;
            }
        }
        return head;
    }
    
    ListNode* swapPairs2(ListNode* head) {
        if (head == NULL || head->next == NULL) return head;
        ListNode* f = head->next;
        ListNode* p1 = head;
        ListNode* p2 = p1->next;
        ListNode* pre = NULL;
        while (p2 != NULL) {
            ListNode* n = p2->next;
            p2->next = p1;
            p1->next = n;
            if (pre != NULL) {
                pre->next = p2;
            }
            pre = p1;
            p1 = n;
            p2 = NULL;
            if (p1 != NULL) {
                p2 = p1->next;
            }
        }
        if (p1 != NULL) {
            pre->next = p1;
            p1->next = NULL;
        }
        return f;
    }
};