LeetCode代码记录 24 Swap Nodes in Pairs

最新推荐文章于 2023-04-23 14:35:52 发布

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本文介绍了一种链表中相邻节点互换位置的算法实现，通过改变节点指针而非节点值来完成交换，确保了常数级别的空间复杂度。

题目如下：

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.

这道题主要还是链表的互换，比较简单，答案运行官方Case时间为4ms，仍有很多提升空间。

/**
 * 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* nextPointer = head;
        ListNode* pPre = NULL;
        ListNode* pSwap = NULL;
        while(nextPointer!=NULL && nextPointer->next!=NULL) {
            if( nextPointer == head) {
                pSwap = nextPointer->next;
                nextPointer->next = pSwap->next;
                head = pSwap;
                pSwap->next = nextPointer;
            } else {
                pSwap = nextPointer->next;
                nextPointer->next = pSwap->next;
                pPre->next = pSwap;
                pSwap->next = nextPointer;
            }
            pPre = nextPointer;
            nextPointer = nextPointer->next;
        }
        return head;
    }
};