LeetCode OJ：Partition List（分割链表）

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.

将链表分割成两部分，大于某个数字的在左侧，小于等于某个数字的在右侧，用的方法比较蠢可能就是遍历一次分成两个链表，然后再将它们接起来。具体代码如下所示：

 class Solution {
 public:
     ListNode* partition(ListNode* head, int x) {
           ListNode * helper1 = new ListNode(INT_MIN);
           ListNode * helper2 = new ListNode(INT_MIN);
           ListNode * p1 = helper1;
           ListNode * p2 = helper2;
           while(head){
               if(head->val < x){
                   p1->next = head;
                   head = head->next;
                   p1 = p1->next;
                   p1->next = NULL;
               }else{
                   p2->next = head;
                   head = head->next;
                   p2 = p2->next;
                   p2->next = NULL;
               }
           }
           p1->next = helper2->next;
           return helper1->next;
     }
 };