【leetcode】Swap Nodes in Pairs （middle）

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.

//start time = 9:54
//end time = 10:16
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

 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;
        //头指针转换
        ListNode * newhead = head->next;
        head->next = newhead->next;
        newhead->next = head;

        ListNode * pre = newhead->next; //之前转换完的最后一个
        ListNode * cur = NULL; //一对中的前一个
        ListNode * next = NULL;//一对中的后一个
        while(pre->next != NULL && pre->next->next != NULL)
        {
            cur = pre->next;
            next = cur->next;
            pre->next = next;
            cur->next = next->next;
            next->next = cur;

            pre = cur;
        }

        return newhead;
    }
};

发现多出了c的选项 用c写的时候 ListNode前面要加 struct 修饰 而C++不用 注意区别 时间上C需要1ms 而C++需要6ms

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *swapPairs(struct ListNode *head) {
    if(head == NULL || head->next == NULL) return head;
        //头指针转换
        struct ListNode * newhead = head->next;
        head->next = newhead->next;
        newhead->next = head;

        struct ListNode * pre = newhead->next; //之前转换完的最后一个
        struct ListNode * cur = NULL; //一对中的前一个
        struct ListNode * next = NULL;//一对中的后一个
        while(pre->next != NULL && pre->next->next != NULL)
        {
            cur = pre->next;
            next = cur->next;
            pre->next = next;
            cur->next = next->next;
            next->next = cur;

            pre = cur;
        }

        return newhead;
}