【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(0), small(0);

        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;

    }

};