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* leftHead = new ListNode(-1); ListNode* rightHead = new ListNode(-1); ListNode* p1 = leftHead; ListNode* p2 = rightHead; ListNode* p = head; while(p){ if(p->val<x){ p1->next = p; p1 = p; p = p->next; }else{ p2->next = p; p2 = p; p = p->next; } } p1->next = rightHead->next; p2->next = nullptr; return leftHead->next; } };
原文：http://www.cnblogs.com/wxquare/p/5928083.html