Swap Nodes in Pairs

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.

这里贴一个错误代码,本人的:

/** * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if(head|| head->next)            return head;        ListNode *p, *q, *swap;        p = head;        q = p->next;        swap = q;        while(p&&q)        {            p->next = q->next;            q->next = p;            p       = p->next;            q       = p->next;        }        return swap;    }};

Submission Result: Runtime Error

Last executed input:{1,2}

原因:

q = p->next; //此时p=NULL了

于是添加　if(p)于前

又错误:

Submission Result: Runtime Error

Last executed input:{}

还是那个原因

if(head||head->next)

head=NULL时,　head->next就错了

改之:

if(!head||!head->next) return head;

当一切就绪时,

/** * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if(!head||!head->next)            return head;        ListNode *p, *q, *swap;        p = head;        q = p->next;        swap = q;        while(p&&q)        {            p->next = q->next;            q->next = p;            p       = p->next;            if(!p)                break;            else                q   = p->next;        }       return swap;    }};

Input:{1,2,3,4}Output:{2,1,3}Expected:{2,1,4,3}

这个错误是,　到后面,　前断开的没连上

增加前面的pre

最终代码:

/** * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if(!head||!head->next)            return head;        ListNode *p, *q, *swap, *pre;        p = head;        q = p->next;        swap = q;        pre  = NULL;        while(p&&q)        {            p->next = q->next;            q->next = p;            if(pre)                pre->next = q;            pre     = p;            p       = p->next;            if(!p)                break;            else                q   = p->next;        }       return swap;    }};