[C语言][LeetCode][328]Odd Even Linked List

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本文介绍了一种算法，用于将单链表中的奇数位置节点与偶数位置节点进行重新排列，确保所有奇数位置节点位于偶数位置节点之前，同时保持原有相对顺序不变。该算法采用两个临时指针遍历链表并重构，实现了O(1)空间复杂度和O(nodes)时间复杂度。

题目

Odd Even Linked List
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 …

难度

简单

分析

题目意思是给定一个单链表，将节点顺序为奇数的放到节点顺序为偶数的前面，形成一个新的链表。解题思路是利用两个临时的指针p、q去遍历原来的链表，拿到对应的奇偶的节点，然后再合并成新的链表。

C代码实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* oddEvenList(struct ListNode* head) {
    struct ListNode *p, *q, *t, *evenHead;
    int count = 1;

    if(!head || !head->next || !head->next->next)
        return head;

    p = head;
    q = head->next;
    evenHead = head->next;
    t = head->next->next;

    while(t)
    {
        if(1 == count)
        {
            count = 2;
            p->next = t;
            p = t;
        }
        else if(2 == count)
        {
            count = 1;
            q->next = t;
            q = t;
        }
        t = t->next;
    }

    q->next = NULL;
    p->next = evenHead;

    return head;    
}