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.
Example
Given 1->4->3->2->5->2->null and x = 3,
return 1->2->2->4->3->5->null.
题解：
Solution 1 ()
class Solution { public: ListNode *partition(ListNode *head, int x) { ListNode* left = new ListNode(-1); ListNode* right = new ListNode(-1); ListNode* l = left; ListNode* r = right; while (head) { if (head->val < x) { l->next = head; l = l->next; } else { r->next = head; r = r->next; } head = head->next; } l->next = right->next; r->next = nullptr; return left->next; } };
原文：http://www.cnblogs.com/Atanisi/p/6847993.html