## Question

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`

.

## Solution

**Result:** Accepted **Time:** 0 ms

Here should be some explanations.

```
struct ListNode* partition(struct ListNode* head, int x) {
struct ListNode low,up;
struct ListNode *llast = &low, *ulast = & up,*tmp;
while(head)
{
tmp = head;
head = head ->next;
if(tmp->val < x)
{
llast -> next = tmp;
llast = tmp;
}
else
{
ulast -> next = tmp;
ulast = tmp;
}
}
ulast->next = NULL;
llast->next = up.next;
return low.next;
}
```

**Complexity Analytics**

- Time Complexity:
- Space Complexity: