[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:
/*algorithm two pointers
create a list l to store the value < x
scan the list, remove those element < x to l
link l with old list
time O(n) space O(1)
*/
ListNode* partition(ListNode* head, int x) {
ListNode dummy(0),*p,*prev,*l,*tail;
dummy.next = head;
prev = &dummy,p = head;
l = tail = NULL;
while(p){
if(p->val < x){
prev->next = p->next;
if(!l)l = tail = p;
else {
tail->next = p;
tail = tail->next;
}
p = prev->next;
}else{
prev = p;
p = p->next;
}
}
if(l)tail->next = dummy.next;
else l = dummy.next;
return l;
}
};