Leetcode 328. Odd Even Linked List

Tyan

发布于 2019-05-25 22:47:33

3280

发布于 2019-05-25 22:47:33

文章被收录于专栏：SnailTyan

文章作者：Tyan

博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

Version 1

/**
 * 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 || !head->next) {
            return head;
        }
        ListNode* even = head->next;
        ListNode* current = head;
        ListNode* next = nullptr;
        while(current->next) {
            next = current->next;
            current->next = current->next->next;
            current = next;
        }
        current = head;
        while(current->next) {
            current = current->next;
        }
        current->next = even;
        return head;
    }
};

Version 2

/**
 * 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 || !head->next) {
            return head;
        }
        int count = 1;
        ListNode* odd = head;
        ListNode* even = head->next;
        ListNode* current = head;
        ListNode* next = nullptr;
        while(current->next) {
            next = current->next;
            current->next = current->next->next;
            current = next;
            count++;
            if(count % 2) {
                odd = current;
            }
        }
        current = head;
        odd->next = even;
        return head;
    }
};

Version 3

/**
 * 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 || !head->next) {
            return head;
        }
        ListNode* odd = head;
        ListNode* even = head->next;
        ListNode* even_head = head->next;
        while(odd->next && even->next) {
            odd->next = odd->next->next;
            even->next = even->next->next;
            odd = odd->next;
            even = even->next;
        }
        odd->next = even_head;
        return head;
    }
};