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：先分成大于等于x 和 小于x 两个链表 再连起来 还是用伪头部
class Solution { public: ListNode *partition(ListNode *head, int x) { ListNode large(0), small(0); ListNode * l = &large; ListNode * s = &small; while(head != NULL) { if(head->val < x) { s = s->next = head; } else { l = l->next = head; } head = head->next; } l->next = NULL; s->next = large.next; return small.next; } };
【leetcode】Partition List（middle）
原文：http://www.cnblogs.com/dplearning/p/4333410.html