leetcode之Partition List

原题如下：

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

此题有两种思路：一种是原地插入，这种思路的关键是找到第一个比x小的节点和第一个比x大的节点，然后一次遍历插入，链表中节点的删除和插入都需要多加小心，另外还需要对特殊情况进行处理。

 ListNode *partition(ListNode *head, int x) {
		if(head == NULL || head->next == NULL)
			return head;
		ListNode*smallRear,*bigHead,*preCur,*cur;		
		if(head->val < x){  //如果head结点比x小，则head即为当前smallRear，然后寻找bigHead
			smallRear = head;
			cur = head->next;
			while(cur != NULL){
				if(cur->val >= x){
				    break;
				}
				cur = cur->next;
				smallRear = smallRear->next;
			}
			if(cur == NULL)//所有元素都比x小，直接返回
			{
			    return head;
			}
			bigHead = cur;
			preCur = cur;
			cur = cur->next;  
		}
		else  //head结点不比x小，则head即为bigHead，需要往后寻找smallRear
		{
			bigHead = head;
			preCur = head;
			cur = head->next;
			while(cur != NULL){
				if(cur->val < x){
				    break;
				}
				preCur = cur;
				cur = cur->next;					
			}
			if(cur == NULL )//所有节点都不小于x，直接返回
			{
				return head;
			}
			preCur->next = cur->next;
			cur->next = bigHead;
			smallRear = cur;
			head = cur;
			cur = preCur->next;
		}
		while(cur != NULL){
			if(cur->val < x){
				preCur ->next = cur->next;
				smallRear->next = cur;
				cur->next = bigHead;
				smallRear = cur;
				cur = preCur->next;
			}
			else
			{
				preCur = cur;
				cur = cur->next;
			}
			
		}
		return head;
    }

另外一种思路则比较简单，创建两个头结点，分别用来链接比x小的元素和比x大的元素，然后将两个链表进行连接即可，在编码时要注意将最后一个比x大的元素的next指针置空，否则有可能出现环。

ListNode *partition(ListNode *head, int x) {
		if(head == NULL || head->next == NULL)
			return head;
		ListNode *p1 = new ListNode(0);
		ListNode *p2 = new ListNode(0);
		ListNode *pSmall = p1;
		ListNode *pBig = p2;
		while(head != NULL){
			if(head->val < x){
				pSmall->next = head;
				pSmall = pSmall->next;
			}
			else
			{
				pBig->next = head;
				pBig = pBig->next;
			}
			head = head->next;
		}		
		pBig->next = NULL;
		pSmall ->next = p2->next;
		head = p1->next;
		delete p1,p2;
		return head;			
	}

两种方法对比发现，第一种思路没有申请额外的空间，但对链表的删除和插入操作较多，需要注意的细节问题较多（我也是经过三次提交才AC），而第二种方法则比较直观易懂，而且不容易出错，另外，链表头结点的存在确实方便了操作。