LeetCode Swap Nodes in Pairs

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        ListNode *a = NULL;
        ListNode *b = NULL;
        ListNode *tail = NULL;
        ListNode *nhead = NULL;
        a = head;
        while (a != NULL && (b = a->next) != NULL) {
            a->next = b->next;
            b->next = a;
            if (tail != NULL) {
                tail->next = b;
            } else {
                nhead = b;
            }
            tail = a;
            a = a->next;
        }
        if (nhead == NULL) {
            nhead = head;
        }
        return nhead;
    }
};

准备六级,水一发

第二轮:

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.

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        if (head == NULL || head->next == NULL) {
            return head;
        } 
        ListNode holder(0);
        ListNode* prev = &holder;
        ListNode* cur = head;
        
        while (cur != NULL) {
            ListNode* next = cur->next;
            if (next == NULL) {
                break;
            }
            head = next->next;
            next->next = cur;
            cur->next = head;
            prev->next = next;
            prev = cur;
            cur = cur->next;
        }
        return holder.next;
    }
};
原文地址:https://www.cnblogs.com/lailailai/p/3766541.html