【一天一道LeetCode】#328 Odd Even Linked List-优快云博客

本文提供了解决LeetCode中关于链表的奇偶节点分组问题的详细步骤，包括代码实现及解析，适用于链表操作的学习者。

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一天一道LeetCode系列

（一）题目

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 …

Credits: Special thanks to @DjangoUnchained for adding this problem
and creating all test cases.

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题目的意思大概是：根据数字编号，将奇数编号排在偶数编号之前，不能改变奇数编号和偶数编号原有的内部顺序。

思路：可以建立两个链表，奇数编号链表和偶数编号链表，遍历原链表，进行判断分类即可。

（二）代码实现

/**
 * 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; //原链表为空，即返回
        if(head->next == NULL) return head;//原链表只有两位，即返回
        ListNode* oddtail = head; //奇数编号链表尾
        ListNode* even = head->next; //偶数编号链表开始位
        ListNode* eventail = head->next;//偶数编号链表尾
        ListNode* p = head->next->next;
        int i;
        while(p){
            if((i & 0x1) == 1)//判断编号为奇数
            {
                oddtail->next = p; 
                oddtail =  oddtail->next;
            }
            else {
                eventail->next = p;
                eventail=eventail->next;
            }
            p=p->next;
            i++;
        }
        eventail-> next = NULL; //链表尾置空
        oddtail->next = even;//将两个链表尾首相连，得到结果
        return head;
    }

};