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：可以将小于x的node放在一个链表中，大于等于x的node放在另一个链表中，然后将两个链表首尾相连。
class Solution { public: ListNode *partition(ListNode *head, int x) { if(head == NULL) return head; ListNode *shead = new ListNode(-1); ListNode *bhead = new ListNode(-1); ListNode *ps = shead, *pb = bhead; while(head){ ListNode *tmp = head; head = head->next; tmp->next = NULL; if(tmp->val < x){ ps->next = tmp; ps = ps->next; }else{ pb->next = tmp; pb = pb->next; } } ps->next = bhead->next; return shead->next; } };
原文：http://www.cnblogs.com/Kai-Xing/p/4109027.html