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。