leetcode链表--16、swap-nodes-in-pairs（成对交换链表结点）

题目描述

 

Given a linked list, swap every two adjacent nodes and return its head.

For example,

Given1->2->3->4, you should return the list as2->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.

 

解题思路：与15题一样，只是k为2，也是使用栈结构进行存储。

1）当tmp<k时，将p插入到栈中，tmp==k是弹出栈，存入新链表中

2）如果最后栈中还有元素，即剩一个，则将其存入令一个栈中，然后将那个栈弹出存入新链表中

3）返回head->next

/**
 * 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)
            return NULL;
        int k = 2;
        stack<ListNode *> s;
        ListNode *p = head;
        ListNode *newHead = new ListNode(0);
        ListNode *newP = newHead;
        int tmp = 0;
        while(p != NULL)
        {
            if(tmp<k)
            {
                s.push(p);
                p = p->next;
                tmp++;
            }
            if(tmp == k)
            {
                while(!s.empty())
                {
                    newP->next = s.top();
                    s.pop();
                    newP = newP->next;
                    tmp--;
                }
            }
 
        }
        stack<ListNode *>s2;
        while(!s.empty())
        {
            s2.push(s.top());
            s.pop();
        }
        while(!s2.empty())
        {
            newP->next = s2.top();
            s2.pop();
            newP = newP->next;
        }
        newP->next = NULL;
        return newHead->next;
    }
};