LeetCode #328 Odd Even Linked List 奇偶链表

328 Odd Even Linked List 奇偶链表

Description:
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:

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL

Constraints:

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 ...
The length of the linked list is between [0, 10^4].

题目描述:
给定一个单链表，把所有的奇数节点和偶数节点分别排在一起。请注意，这里的奇数节点和偶数节点指的是节点编号的奇偶性，而不是节点的值的奇偶性。

请尝试使用原地算法完成。你的算法的空间复杂度应为 O(1)，时间复杂度应为 O(nodes)，nodes 为节点总数。

示例 :

示例 1:

输入: 1->2->3->4->5->NULL
输出: 1->3->5->2->4->NULL

示例 2:

输入: 2->1->3->5->6->4->7->NULL
输出: 2->3->6->7->1->5->4->NULL

说明:

应当保持奇数节点和偶数节点的相对顺序。
链表的第一个节点视为奇数节点，第二个节点视为偶数节点，以此类推。

思路:

双指针分别指向奇数节点和偶数节点
分别遍历赋值
最后将奇数和偶数链表连接起来
时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

/**
 * 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) return head;
        ListNode *p = head, *q = head -> next, *r = head -> next;
        while (q and q -> next) 
        {
            p -> next = p -> next -> next;
            q -> next = q -> next -> next;
            p = p -> next;
            q = q -> next;
        }
        p -> next = r;
        return head;
    }
};

Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null) return head;
        ListNode p = head, q = head.next, r = head.next;
        while (q != null && q.next != null) {
            p.next = p.next.next;
            q.next = q.next.next;
            p = p.next;
            q = q.next;
        }
        p.next = r;
        return head;
    }
}

Python:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if not head:
            return head
        p, q, r = head, head.next, head.next
        while q and q.next:
            p.next, q.next, p, q = p.next.next, q.next.next, p.next.next, q.next.next
        p.next = r
        return head