Swap Nodes in Pairs

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.

分析： 比较简单的链表操作 需要将a->b->(b->next) 转换为b->a->(b->next)这样的形式

自己写的代码为

/**
 * 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) {
        if(head ==nullptr || head->next ==nullptr)
            return head;
        ListNode* cur = head;
        ListNode* curNext = cur->next;
        cur->next = curNext->next;
        curNext->next = cur;
        head= curNext;
        ListNode* prev = cur;
        while(prev && prev->next){
            cur = prev->next;
            curNext = cur->next;
            if(curNext){
                 cur->next = curNext->next;
                 curNext->next = cur;
                 prev->next = curNext;
                 prev = cur;
            }else{
                cur->next =curNext;
                prev = nullptr;
            }
            
        }
        return head;
    }
};

Discuss中提出的用指针的指针解决方案，稍微难一点，但很高效

ListNode* swapPairs(ListNode* head) {
    ListNode **pp = &head, *a, *b;
    while ((a = *pp) && (b = a->next)) {
        a->next = b->next;
        b->next = a;
        *pp = b;
        pp = &(a->next);
    }
    return head;
}