leetcode_328_Odd Even Linked List(easy)(C++)

最新推荐文章于 2021-11-23 20:28:38 发布

Lan_MuLan

最新推荐文章于 2021-11-23 20:28:38 发布

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文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/u010431315/article/details/50969466

本文介绍了一种在单链表中将所有奇数位置节点聚集在一起，随后连接所有偶数位置节点的方法。该方法实现了原地修改，空间复杂度为O(1)，时间复杂度为O(n)。通过使用两个指针分别跟踪已排列好的奇数和偶数节点的末尾，实现了链表的正确重组。

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Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

Note:
The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on ...

思路：两个指针，第一个指针指向已排好链表的基数最后一个，第二个指针指向已排好链表的偶数最后一个。另外用另一指针指向偶数第一个。

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* oddEvenList(ListNode* head) {
if (!head || !head->next) return head;
ListNode *pre = head, *cur = head->next;
while (cur && cur -> next) {
ListNode *tmp = pre->next;
pre->next = cur->next;
pre = pre->next;
cur->next = pre->next;
pre->next = tmp;
cur = cur->next;
}
return head;
}
};