86. Partition List (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
.
class Solution { public: ListNode *partition(ListNode *head, int x) { if(!head || !(head->next) ) return head; ListNode *current = head; ListNode *smallPointer = NULL; //point to the last node <x ListNode *largePointer = NULL; //point to the last node >x while(current) { if(current->val >= x) { largePointer = current; current = current->next; } else { if(!largePointer) { smallPointer = current; current = current->next; } else if(smallPointer) { largePointer->next = smallPointer->next; smallPointer -> next = current; current = current->next; smallPointer = smallPointer->next; smallPointer->next = largePointer->next; largePointer->next = current; } else //head { smallPointer = current; current = current->next; smallPointer->next = head; head = smallPointer; largePointer->next = current; } } } return head; } };