LeetCode75——Day30

本文主要是介绍LeetCode75——Day30，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

一、题目

328. Odd Even Linked List

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

Example 1:

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]
Example 2:

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

Constraints:

The number of nodes in the linked list is in the range [0, 104].
-106 <= Node.val <= 106

二、题解

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* oddEvenList(ListNode* head) {if(head == nullptr) return nullptr;ListNode* evenHead = head->next;//奇指针ListNode* odd = head;//偶指针ListNode* even = evenHead;while(even && even->next){odd->next = even->next;//奇指针向后移动odd = odd->next;even->next = odd->next;//偶指针向后移动even = even->next;}odd->next = evenHead;return head;}
};

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