LeetCode 86. Partition List 划分链表 C++

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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

有点用到了倒置链表II的方法，将符合要求的结点放置在指向pre指针的后面。这道题的思路应该是找到第一个大于等于x值的结点，他前一个位置始终定位pre指针，放置比x小的结点。

方法一（C++）

 class Solution {
 public:
     ListNode* partition(ListNode* head, int x) {
         ListNode* dummy=new ListNode(-);
         dummy->next=head;
         ListNode* pre=dummy,* cur=head;
         while(pre->next&&pre->next->val<x)
             pre=pre->next;
         cur=pre;
         while(cur->next){
             if(cur->next->val<x){
                 ListNode* t=cur->next;
                 cur->next=t->next;
                 t->next=pre->next;
                 pre->next=t;
                 pre=t;
             }
             else
                 cur=cur->next;
         }
         return dummy->next;
     }
 };