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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.

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;	}};

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