58-Odd Even Linked List

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最新推荐文章于 2025-09-05 22:51:45 发布

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分类专栏： Coding-Leetcode等文章标签： odd even linked list

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本文介绍了一种在O(1)空间复杂度和O(n)时间复杂度下，对单链表进行奇偶节点重组的方法。该算法通过调整指针指向实现原地重排，保持了各组内部节点的原始顺序。

Odd Even Linked List My Submissions QuestionEditorial Solution
Total Accepted: 29496 Total Submissions: 77710 Difficulty: Medium
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：主要是操作指针指向
时间：O(n) 节点数n
空间：O(1)

/**
 * 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||head->next==NULL)return head;//头部为空或1个
        ListNode * oddhead = head,*evenhead=head->next;
        ListNode *op=oddhead,*ep=evenhead,*opend=oddhead;//奇偶链首节点
        while(op!=NULL&&ep!=NULL){  //循环修改指向
            op->next = ep->next;
            if(op->next!=NULL)ep->next = op->next->next;
            opend = op;
            if(op!=NULL)op = op->next;
            if(ep!=NULL)ep = ep->next;
        }
        if(op!=NULL)opend=op;
        opend->next = evenhead;
        return head;
    }
};