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.

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