LeetCode—Reorder List

Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}

.

想法比较简单

利用快慢两个指针将链表一分为二，针对第二个子链表求倒序，最后将两个子链表合并。

一分为二的方法参考了前面链表排序中用到的归并排序用到的算法

这里代码写得比较长，但是好像在网上也没有找到更好的更简洁的，如果有更好的方法欢迎交流

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
	void reorderList(ListNode *head) {
		if(!head || !head->next)
		{
			return ;
		}
ListNode * midL = head;
		ListNode * midR = head;
		ListNode * midPre = NULL;
		while(midR && midR->next)  //<find the middle point
		{
			midR = midR->next->next;
			midL= midL->next;
		}
		
		ListNode *pre  = midL->next;
		midL->next = NULL;
		ListNode *pCur = NULL;
		
		if (pre)   //<reverse the second part of the link
		{
			pCur = pre;
			if(pre->next)
			{
				pCur = pre->next;
				pre->next = NULL;  //<the last point in the reverse list should point to null
				while(pCur)
				{
					ListNode *pNext = pCur->next;
					if(pNext)
					{
						pCur->next = pre;
						pre = pCur;
						pCur = pNext;
					}
					else
					{
						pCur->next = pre;
						break;
					}
				}
			}
		}
		
		ListNode * q = head;           //merge the two part
		while(q && pCur)
		{
			ListNode *tempL = q->next;
			ListNode *tempR = pCur->next;
			q->next = pCur;
			pCur->next = tempL;
			pCur = tempR;
			q = tempL;
		}
	//	return head;
	}
};