328.Odd Even Linked List

Tags: Medium Linked List

Links: https://leetcode.com/problems/odd-even-linked-list/

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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 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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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!head || !head -> next) return head;
        ListNode *dummy = new ListNode(-1);
        dummy -> next = head;
        ListNode *odd = head, *even = head -> next;
        ListNode *pre = even;
        while (even && even -> next) {
            odd -> next = even -> next;
            even -> next = odd -> next -> next;
            odd -> next -> next = pre;  
            odd = odd -> next;
            even = even -> next;
        }

        ListNode *res = dummy -> next;
        delete dummy; dummy = NULL;
        return res;
    }
};

链表的问题画个图就明白了，然后就知道该用多少个指针了。一般考虑链表节点个数为0，1，2，3，4，如果都正确，那么后面的也就都正确了。可以自行验证。