Leetcode 86 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

.
给链表，将小于X值得搬到前面
用两个链连接两种类型的结点，最后把大链加到小链后面，

注意把最后一位的指针断开，否则可能形成循环链表。

/**
* 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) {
ListNode* small=new ListNode(0);
ListNode* result=small;
ListNode* big=new ListNode(0);
ListNode* result2=big;
while(head!=NULL)
{
if(head->val>=x)
{
big->next=head;
big=big->next;
}
else
{

small->next=head;
small=small->next;
}
head=head->next;
}
big->next=NULL;//把最后一位指针断开，否则会超时
small->next=result2->next;//合并
return result->next;
}
};