LeetCode 24: Swap Nodes in Pairs

最新推荐文章于 2019-11-27 22:42:08 发布

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本文介绍了一种链表操作算法，实现每两个相邻节点进行交换。通过定义链表节点结构，采用迭代方式遍历链表，实现了节点的交换，并在不修改节点值的情况下仅改变节点链接关系。

Swap Nodes in Pairs

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.

解题思路

代码如下：

/**
 * 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) {
        if (head == NULL || head->next == NULL) return head;

        ListNode *pre = NULL, *p1 = head, *p2 = head->next;
        head = head->next;
        while (p1 && p2) {
            p1->next = p2->next;
            p2->next = p1;
            if (pre != NULL) {
                pre->next = p2;
            }
            pre = p1;

            p1 = p1->next;
            if (p1 != NULL)
                p2 = p1->next;
        }

        return head;
    }
};