[LeetCode]88. Swap Nodes in Pairs链表成对逆序

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.

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解法：题目要求不能修改节点的值，因此只能调整节点的指针。每次调整两个节点，依次往前移动。

/**
 * 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 == NULL || head->next == NULL) return head;
        ListNode* help = new ListNode(0);
        help->next = head;
        ListNode *first = help, *second = head, *third = head->next;
        while (third != NULL) {
            ListNode* tmp = third->next;
            third->next = first->next; // 调整一对节点
            first->next = third;
            second->next = tmp;
            if (tmp == NULL) break;
            first = second; // 前移到下一对节点
            second = tmp;
            third = tmp->next;
        }
        return help->next;
    }
};