86. Partition List

题目：

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：

处理边界情况；将链表按x值拆为两段，再合并起来就好了。（感觉难度不大，可能我的方法不好）

代码：

/** * 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) {        if (head == nullptr || head->next == nullptr)   return head;        ListNode* phead1 = new ListNode(0);        ListNode* phead2 = new ListNode(0);        ListNode* p1 = phead1;        ListNode* p2 = phead2;        while (head){            if (head->val < x){                p1->next = head;                head = head->next;                p1 = p1->next;            }            else{                p2->next = head;                head = head->next;                p2 = p2->next;            }        }        p2->next = nullptr;        p1->next = phead2->next;        delete phead2;        head = phead1->next;        delete phead1;        return head;    }};