[Leetcode] Partition List

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

.

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(!head) return NULL;

ListNode *smallVir = new ListNode(-1);
ListNode *greatVir = new ListNode(-1);

ListNode *smallPtr = smallVir; //Virtual head for smaller list
ListNode *greatPtr = greatVir;
ListNode *curPtr   = head;

while(curPtr!=NULL)
{

if(curPtr->val<x)
{
smallPtr->next = curPtr;
smallPtr = curPtr;
}
else
{
greatPtr->next= curPtr;
greatPtr = curPtr;
}

curPtr = curPtr->next;

}

greatPtr->next = NULL;//重要，新链表的末尾不一定是旧链表的末尾
smallPtr->next = (greatVir->next)?greatVir->next:NULL;
return smallVir->next;
}
};