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

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;
    }
};