[Leetcode] 328. Odd Even Linked List 解题报告

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本文介绍了一种在O(1)空间复杂度和O(n)时间复杂度下，对链表进行奇偶节点分组的方法。该方法通过三个步骤实现：首先分离奇数和偶数位置的节点；其次寻找奇数链表的尾节点；最后将偶数链表连接到奇数链表的尾部。

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 ...

思路：

三个步骤：1）把奇数位和偶数位上的结点分别都链接起来；2）找到奇数位的新链表的末尾结点；3）将偶数位的新链表接到奇数位的新链表的末尾结点上。大功告成！时间复杂度是O(n)，空间复杂度O(1)，n是输入链表的长度。

代码：

/**
 * 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 || head->next == NULL) {
            return head;
        }        
        ListNode* oddHead = head;
        ListNode* evenHead = head->next;
        ListNode* node = head;
        while(node != NULL && node->next != NULL) {     // link all the odd/even node together
            ListNode* nextNode = node->next;
            node->next = nextNode->next;
            node = nextNode;
        }
        node = oddHead;
        while(node->next != NULL) {                     // find the last node in the odd linkedlist
            node = node->next;
        }
        node->next = evenHead;                          // join the even linkedlist at the end of the odd linkedlist
        return head;
    }
};