(链表) leetcode 328. Odd Even Linked List

佚名 6年前 (2019-04-14) 随笔 1550人围观抢沙发百度已收录

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.

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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

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 ...

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这个题是需要建立两个辅助结点，分别指向奇数位置和偶数位置。最好画图来辅助思考。

C++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) {
        ListNode *odd,*even;
        if(!head || !head->next) return head;
        odd = head;
        even = head->next;
        while(even && even->next){
            ListNode *cur = odd->next;   //十分有必要建立这个辅助结点。
            odd->next = even->next;
            even->next = even->next->next;
            odd->next->next = cur;
            even = even->next;
            odd = odd->next;
        }
        return head;
    }
};