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社区首页 >专栏 >Leetcode 328. Odd Even Linked List

Leetcode 328. Odd Even Linked List

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Tyan
发布2019-05-25 22:47:33
3280
发布2019-05-25 22:47:33
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文章被收录于专栏:SnailTyan

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文章作者:Tyan

博客:noahsnail.com | CSDN | 简书

1. Description

2. Solution

  • Version 1
代码语言:javascript
复制
/**
 * 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
代码语言:javascript
复制
/**
 * 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
代码语言:javascript
复制
/**
 * 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;
    }
};

Reference

  1. https://leetcode.com/problems/odd-even-linked-list/description/
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原始发表:2018年09月29日,如有侵权请联系 cloudcommunity@tencent.com 删除

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目录
  • 1. Description
  • 2. Solution
  • Reference
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