标签：
题目：
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.
代码：
/** * 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 dummyLess(-1), dummyGreaterOrEqual(-1); ListNode *p1 = &dummyLess, *p2 = &dummyGreaterOrEqual, *p = head; while(p){ if ( p->val < x){ p1->next = p; p1 = p1->next; } else{ p2->next = p; p2 = p2->next; } p = p->next; } p1->next = dummyGreaterOrEqual.next; p2->next = NULL; return dummyLess.next; } };
Tips:
链表的基础操作。
设立两个虚表头（代表两个子链表），分别往后接小于x的和大于等于x的，然后再把两个子链表接起来。
标签：
原文地址：http://www.cnblogs.com/xbf9xbf/p/4465000.html
