Swap Nodes in Pairs

最新推荐文章于 2019-05-11 21:02:45 发布

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最新推荐文章于 2019-05-11 21:02:45 发布

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原文链接：http://www.cnblogs.com/pengyu2003/p/3572417.html

本文介绍了一种链表操作算法，重点在于如何在不修改节点值的情况下交换相邻两个节点的位置，通过逐步模拟实现链表节点间的高效互换。

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.

代码不是最简的，处理head的部分明显可以和后面合并。关键记住无论思路多清晰都需要模拟一下。

/**
 * 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) {
        ListNode *first  = head;
        
        if(head == NULL) return NULL;
        ListNode *second = head -> next;
        ListNode *temp;
        
        if(second != NULL)
        {
             head = second;
             temp = first;
             first -> next = second -> next;
             second -> next = first;
             first = first->next;
             if(first == NULL ||first -> next == NULL )return head;
             second = first ->next;
             temp ->next = second;
             
        }
        
        while(first != NULL && second != NULL)
        {
             temp = first;
             first -> next = second -> next;
             second -> next = first;
             first = first->next;
             if(first == NULL ||first -> next == NULL )return head;
             second = first ->next;
             temp ->next = second;
        }
        
        return head;
        
    }
};

转载于:https://www.cnblogs.com/pengyu2003/p/3572417.html