34：Swap Nodes in Pairs

最新推荐文章于 2019-10-13 16:37:25 发布

原创最新推荐文章于 2019-10-13 16:37:25 发布 · 197 阅读

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本文提供了一种解决方案，用于交换链表中每两个相邻节点的位置，同时确保仅使用常数空间复杂度。给出的示例代码展示了如何在不改变节点值的情况下实现这一目标。

题目：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.

解题代码如下：

class Solution {
public:
        ListNode* swapPairs(ListNode* head) {
                ListNode dummy{-1};
                dummy.next = head;
                ListNode *prev = &dummy, *cur = dummy;
                for (int i = 0; cur && i != 2; ++i)
                        cur = cur -> next;
                if (cur == 0) return head;
                for (int i = 0; ; ++i) {
                        if (i % 2 == 0) {
                                prev -> next -> next= cur -> next;
                                cur -> next = prev -> next;
                                prev -> next = cur;
                                cur = cur -> next;
                        }
                        if (cur -> next == nullptr) break;
                        cur = cur -> next;
                        prev = prev -> next;
                }
                return dummy.next;
        }
};

还有一个更简洁的代码（代码参考自网址https://github.com/soulmachine/leetcode#leetcode题解），但如下代码不符合题意要求（题意要求不允许修改结点的值）

class Solution {
public:
        ListNode* swapPairs(ListNode* head) {
                ListNode* p = head;
                while (p && p -> next) {
                        swap(p -> val, p -> next -> val);
                        p = p -> next -> next;
                }
                return head;
        }
};