【leetcode】Partition List（middle）

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.

思路：先分成大于等于x 和 小于x 两个链表 再连起来  还是用伪头部

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode large(), small();
        ListNode * l = &large;
        ListNode * s = &small;

        while(head != NULL)
        {
            if(head->val < x)
            {
                s = s->next = head;
            }
            else
            {
                l = l->next = head;
            }
            head = head->next;
        }

        l->next = NULL;
        s->next = large.next;
        return small.next;
    }
};