[LeetCode]Swap Nodes in Pairs

题目：Swap Nodes in Pairs

交换相邻的两个节点。

思路：

两个指针指向相邻的两个节点，交换他们。

注意：

链表节点少于2个直接返回。

1->2->3->4->5->6

p:2,q:3;

交换3,4的顺序：

p->next = q->next;
p = p->next;
q->next = p->next;
p->next = q;

/*******************************************************************************
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.
*******************************************************************************/
#include<stdio.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

/**
 * 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;//少于两个节点

    struct ListNode* p = head->next;//先交换开头两个节点
    head->next = p->next;
    p->next = head;
    head = p;
    p = p->next;
    struct ListNode* q = p->next;
    while(p->next != NULL && q->next != NULL){
        p->next = q->next;
        p = p->next;
        q->next = p->next;
        p->next = q;
        q = q->next;
        p = p->next;
    }

        return head;
}

void display(struct ListNode* p){
    while(p != NULL){
        printf("%d ",p->val);
        p = p->next;
    }
    printf("
");
}

void main(){
    int n = 7;
    struct ListNode* head = NULL;
    for(int i = 0;i < n;i++){
        struct ListNode* p = (struct ListNode* )malloc(sizeof(struct ListNode));
        p->val = i;
        p->next = head;
        head = p;
    }
    display(head);
    head = swapPairs(head);
    
    struct ListNode* p = NULL;
    while(head != NULL){
        p = head;
        head = head->next;
        printf("%d ",p->val);
        free(p);
    }
}