LeetCode 436. Find Right Interval 题解（C++）

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题目描述

• 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.

举例

• Given 1->2->3->4->5->NULL,
  return 1->3->5->2->4->NULL.

补充

• 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

- 具体看上面的图解，OJ上给出的代码是通过两个分别指向奇数结点和偶数结点的指针odd,even来完成的，将奇数结点接在odd链表上，偶数结点接在even链表上，最后再将even链表接在odd链表后面；
- 我的思路跟上面的差不多，差别在于不是在最后才将偶数链表接在奇数链表后面，而是每次都将奇数结点和偶数链表交换位置，详情见代码。

代码

我的代码

class Solution {public:    ListNode* oddEvenList(ListNode* head)    {        if (head == NULL || head->next == NULL)        {            return head;        }        ListNode *preNode = head;        ListNode *curNode = preNode;        ListNode *nextNode = curNode->next;        while (nextNode != NULL)        {            curNode = nextNode;            nextNode = nextNode->next;            if (nextNode == NULL)            {                break;            }            ListNode *temp = nextNode->next;            nextNode->next = preNode->next;            preNode->next = nextNode;            curNode->next = temp;            preNode = preNode->next;            nextNode = curNode->next;        }        return head;    }};

leetcode给出的代码

class Solution {public:    ListNode* oddEvenList(ListNode* head)    {        if (head == NULL)        {            return head;        }        ListNode *odd = head;        ListNode *evenHead = head->next;        ListNode *even = evenHead;        while (even != NULL && even->next != NULL)        {            odd->next = even->next;            odd = odd->next;            even->next = odd->next;            even = even->next;        }        odd->next = evenHead;        return head;    }};