328. Odd Even Linked List

本文介绍了一种在O(1)空间复杂度和O(nodes)时间复杂度下,将单链表中的奇数节点和偶数节点分别组合的方法。通过创建两个虚拟头结点,分别用于连接奇数和偶数节点,并保持原有的相对顺序不变。

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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.

解题思路:

类似86. Partition List

/**
 * 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_dummy(-1);
        ListNode even_dummy(-1);

        ListNode *odd_cur = &odd_dummy;
        ListNode *even_cur = &even_dummy;

        int i = 0;
        for (ListNode *cur = head; cur; cur = cur->next) {
            i++;
            if (i % 2) {
                odd_cur->next = cur;
                odd_cur = cur;
            }
            else {
                even_cur->next = cur;
                even_cur = cur;
            }
        }

        odd_cur->next = even_dummy.next;
        even_cur->next = NULL;

        return odd_dummy.next;
    }
};
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