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

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* partition(ListNode* head, int x) {
12         ListNode *leftH, *leftT, *rightH, *rightT, *p = head;
13         leftH = leftT = new ListNode(0);
14         rightH = rightT = new ListNode(0);
15         
16         while(p != NULL) {
17             if(p->val < x) {
18                 leftT->next = p;
19                 leftT =leftT->next;
20                 
21             }
22             else if(p->val >= x) {
23                 rightT->next = p;
24                 rightT = rightT->next;
25             }
26             p = p->next;
27         }
28         
29         leftT->next = rightH->next;
30         rightT->next = NULL;
31         
32         return leftH->next;
33     }
34 };

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