LeetCode 143 Reorder List Add to List-优快云博客

本文介绍了一种链表重组算法，该算法将给定的单向链表重新排列为指定的交错模式，即首尾元素交替出现。通过计算链表长度并找到中点，将链表分为两部分，接着反转后半部分链表，并将其与前半部分交错连接起来。

题目：

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

想法：将后半段倒序后，插入前半段

代码（c）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void reorderList(struct ListNode* head) {
        struct ListNode* mid_head ;
        struct ListNode* cur_ptr,*pre_ptr;
        int i,length;
        
        mid_head = (struct ListNode *)malloc(sizeof(struct ListNode));
        mid_head->val = 0;
        mid_head->next = NULL;
        
        if(head != NULL)
        {
           length = 0;
            cur_ptr = head;
            while(cur_ptr!=NULL)
            {
                length++;
                cur_ptr = cur_ptr->next;
            }
            for(i = 0, pre_ptr = cur_ptr = head; i<length;i++)
            {
                if(i>length/2&&cur_ptr!=NULL)
                {
                    pre_ptr->next = cur_ptr->next;
                    cur_ptr->next = mid_head->next;
                    mid_head->next = cur_ptr;
                    cur_ptr = pre_ptr->next;
                }
                else
                {
                    pre_ptr = cur_ptr;
                    cur_ptr = cur_ptr->next;
                }
            }
            pre_ptr = head;
            cur_ptr = mid_head->next; 
            while(cur_ptr!=NULL)
            {
                mid_head->next = cur_ptr->next;
                cur_ptr->next = pre_ptr->next;
                pre_ptr->next = cur_ptr;
                pre_ptr = cur_ptr->next;
                cur_ptr = mid_head->next;
            }
        }
}

Runtime：16ms