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;
}