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  • [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;
        }
    };
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  • 原文地址:https://www.cnblogs.com/aprilcheny/p/4969766.html
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