Partition List

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.

设定两个头结点，一个用来连接大于等于x的结点，令一个用于连接小于x的结点，最后将大于的结点挂在小于x的结点的后面。

class Solution {public:ListNode *partition(ListNode *head, int x) {if(!head || !head->next) return head;ListNode *big = new ListNode(1);ListNode *small = new ListNode(-1);ListNode *b = big;ListNode *s = small;ListNode *p = head;while (head){ListNode *pnext = head->next;if(head->val >= x){head->next = b->next;b->next = head;b = head;}else{head->next = s->next;s->next = head;s = head;}head = pnext;}s->next = big->next;head = small->next;delete small;delete big;return head;}};