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 tox.
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) {}
* };
*/
#define LN ListNode
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
LN sguard(-1);
LN hguard(-1);
LN* sTail=&sguard;
LN* hTail=&hguard;
while(head)
{
if (head->val<x)
{
sTail->next=head;
sTail=head;
}
else
{
hTail->next=head;
hTail=head;
}
head=head->next;
}
sTail->next=hguard.next;
hTail->next=NULL;
return sguard.next;
}
};