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) {
if(!head){
return head;
}
ListNode leftDummy(0);
leftDummy.next = NULL;
ListNode rightDummy(0);
rightDummy.next = NULL;
ListNode * curLeft = &leftDummy;
ListNode * curRight = &rightDummy;
ListNode * cur = head;
while(cur){
if(cur->val < x){
curLeft->next = cur;
curLeft = curLeft->next;
}else{
curRight->next = cur;
curRight = curRight->next;
}
cur = cur->next;
}
curLeft->next = NULL;
curRight->next = NULL;
if(!leftDummy.next){
return rightDummy.next;
}else{
curLeft->next = rightDummy.next;
return leftDummy.next;
}
}
};