86. Partition List

Description:

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.

分析：建立两个指针pass1和pass2， head指针遍历链表，小于x的Node链接到pass1上，大于x的Node链接到pass2上，最后把pass2接到pass1之后，返回pass1.（如果只用一个指针，则需要对链表进行删除这样的操作，会更复杂）。

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 new_head1(0), new_head2(0);//建立两个Node来保存两条链表的头结点        ListNode* pass1, *pass2;        pass1 = &new_head1;        pass2 = &new_head2;                while(head)        {            if(head->val < x)            {                pass1->next = head;                pass1 = pass1->next;            }                            else            {                pass2->next = head;                pass2 = pass2->next;            }                              head = head->next;        }                pass2->next = NULL;        pass1->next = new_head2.next;        return new_head1.next;    }};