LeetCode 328. Odd Even Linked List

最新推荐文章于 2025-02-03 15:37:10 发布

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本文介绍了一种在不额外使用空间的情况下对单链表进行奇偶节点分组的方法。通过两个虚拟头节点分别收集奇数位置和偶数位置的节点，并最终将偶数节点链表追加到奇数节点链表之后，实现原地重构。此算法确保了空间复杂度为O(1)，时间复杂度为O(n)。

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 ...

/**
 * 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 == NULL) {
            return head;
        }
        ListNode dummy_odd(0), dummy_even(0);
        ListNode *tail_odd = &dummy_odd;
        ListNode *tail_even = &dummy_even;
        ListNode *cur = head;
        while (cur != NULL) {
            tail_odd->next= cur;
            tail_odd = cur;
            cur = cur->next;
            if (cur != NULL) {
                tail_even->next = cur;
                tail_even = cur;
                cur = cur->next;
            }
            else {
                break;
            }
        }
        tail_even->next = NULL;
        tail_odd->next= dummy_even.next;
        return head;
    }
};