eetcode_c++：链表：Swap Nodes in Pairs(024)

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.

//每一对节点交换

算法

O（N）

#include <iostream>
#include <vector>
#include<cmath>
#include <algorithm>

using namespace std;

struct ListNode{
int val;
ListNode* next;
ListNode(int x):val(x),next(NULL){}
};

//leetcode-------------------
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode* dummy=new ListNode(-1);
dummy->next=head;

ListNode *preNode=dummy,*curNode=head;
int cnt=1;

while(curNode && curNode->next){
preNode->next=curNode->next;
curNode->next=preNode->next->next;
preNode->next->next=curNode;

// go over two nodes
preNode=curNode;
curNode=curNode->next;
}

head=dummy->next;
delete dummy;   // 删除头结点
return head;
}
};
//-------------
//---数组数据插入链表----------------

ListNode* createList(int a[],int n){

ListNode* head=NULL,*p=NULL;
for(int i=0;i<n;i++){
if(head==NULL){
head=p=new ListNode(a[i]);
}else{
p->next=new ListNode(a[i]);
p=p->next;
}
}
return  head;
}

void printList(ListNode* h){
while(h!=NULL){
printf("%d ",h->val);
h=h->next;
}
printf("\n");
}

//-------------

int main(){

Solution s;

int a[]={1,3,5,6,7,10};

ListNode* p1=createList(a,sizeof(a)/sizeof(int));

printList(s.swapPairs(p1));

return 0;
}