LeetCode 分隔链表

题目链接：https://leetcode-cn.com/problems/partition-list/

题目大意

　　略。

分析

　　空间复杂度 O(1) 的做法蛮有意思的，另外加头结点可以少写很多代码。

代码如下

 /**
  * 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) {
         ListNode *leftH, *leftT, *rightH, *rightT, *p = head;
         leftH = leftT = new ListNode();
         rightH = rightT = new ListNode();

         while(p != NULL) {
             if(p->val < x) {
                 leftT->next = p;
                 leftT =leftT->next;

             }
             else if(p->val >= x) {
                 rightT->next = p;
                 rightT = rightT->next;
             }
             p = p->next;
         }

         leftT->next = rightH->next;
         rightT->next = NULL;

         return leftH->next;
     }
 };