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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.
使用两个链表newHead1, newHead2，遍历原链表，如果结点值小于x，则挂在newHead1上，如果大于等于x，则挂在newHead2上，最后把newHead2挂在newHead1上。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* partition(ListNode* head, int x) { 12 ListNode* newHead1 = new ListNode(0); 13 ListNode* newHead2 = new ListNode(0); 14 15 ListNode* pNode1 = newHead1; 16 ListNode* pNode2 = newHead2; 17 18 ListNode* pNode = head; 19 20 while(pNode){ 21 if(pNode->val < x){ 22 pNode1->next = pNode; 23 pNode1 = pNode1->next; 24 }else{ 25 pNode2->next = pNode; 26 pNode2 = pNode2->next; 27 } 28 pNode = pNode->next; 29 } 30 31 pNode2->next = NULL; 32 pNode1->next = newHead2->next; 33 34 return newHead1->next; 35 } 36 };
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原文地址：http://www.cnblogs.com/sankexin/p/5874911.html
