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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.
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/class Solution {public:ListNode* partition(ListNode* head, int x) {ListNode *head1,*head2;head1=head2=NULL;ListNode *cur1,*cur2;cur1=cur2=NULL;while(head){ListNode *cur=head;head=head->next;cur->next=NULL;if(cur->val<x){if(!head1){head1=cur;cur1=head1;}else{cur1->next=cur;cur1=cur1->next;}}else{if(!head2){head2=cur;cur2=head2;}else{cur2->next=cur;cur2=cur2->next;}}}if(!cur1)return head2;cur1->next=head2;return head1;}};
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原文地址：http://www.cnblogs.com/zhoudayang/p/5024432.html
