2.2.3 Partition List

Notes:
	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: ...
	*/
	/**
	* 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) {
        ListNode leftdummy(-1);
        ListNode rightdummy(-1);
        ListNode * lhead = &leftdummy;
        ListNode * rhead = &rightdummy;
        
        for(ListNode*cur = head; cur; cur=cur->next){
            if(cur->val<x){
                lhead->next = cur;
                lhead = lhead->next;
            }else{
                rhead->next = cur;
                rhead = rhead->next;
            }
        }
        lhead->next = rightdummy.next;
        rhead->next = nullptr;
        return leftdummy.next;
    }

    ListNode *partition_1(ListNode *head, int x) {
        ListNode dummy(0), *ins = &dummy, *cur = &dummy;
        dummy.next = head;
        while (cur->next) 
        {
            if (cur->next->val >= x)
            {
                cur = cur->next;
            } 
            else 
            {
                if (cur == ins) 
                {
                    cur = cur->next;
                    ins = ins->next;
                } 
                else 
                {
                    ListNode *move = cur->next;
                    cur->next = move->next;
                    move->next = ins->next;
                    ins->next = move;
                    ins = move;
                }
            }
        }
        return dummy.next;
    }
};