34. Swap Nodes in Pairs

  1. Swap Nodes in Pairs My Submissions QuestionEditorial Solution
    Total Accepted: 95230 Total Submissions: 270562 Difficulty: Easy
    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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思路:简单,学会指针间的操作和指向

/**
 * 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 *beg = head->next,*p=head,*pre=head;
        while(p!=NULL&&p->next!=NULL)
        {
            ListNode *p_next_next = p->next->next;
            pre->next = p->next;
            p->next->next = p;
            p->next = p_next_next;
            pre = p;
            p = p->next;
        }
        return beg;
    }
};
原文地址:https://www.cnblogs.com/freeopen/p/5482935.html