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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.
public ListNode Partition(ListNode head, int x) { if(head == null) return head; ListNode smallStart =new ListNode(0); ListNode largeStart =new ListNode(0); ListNode p1 = smallStart; ListNode p2 = largeStart; //assign small and large pointer while(head != null) { if(head.val < x) { smallStart.next =head; smallStart = smallStart.next; } else { largeStart.next = head; largeStart = largeStart.next; } head = head.next; } largeStart.next = null; smallStart.next = p2.next; return p1.next; }
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原文地址：http://www.cnblogs.com/renyualbert/p/5870486.html
