[Leetcode] 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.

No words!

/**
 * 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 == NULL) return head;
        ListNode *h = new ListNode(-1);
        h->next = head;
        ListNode *pre = h, *cur = h, *tmp;
        while (cur != NULL && cur->next != NULL) {
            if (cur->next->val < x && cur != pre) {
                tmp = cur->next;
                cur->next = tmp->next;
                tmp->next = pre->next;
                pre->next = tmp;
                pre = pre->next;
            } else if (cur->next->val < x && cur == pre) {
                pre = pre->next;
                cur = cur->next;
            } else {
                cur = cur->next;
            }
        }
        return h->next;
    }
};