标签：
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.
分析：用sm、sm_tail、bg、bg_tail分别记录小于x的头、尾、大于等于x的头、尾。
用时：8ms
特别注意注释部分！！在给head->next指定新的位置之前不能将其置空，否则会出问题喔
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 if(!head || !head->next) return head; 13 14 ListNode* sm = new ListNode(0); 15 ListNode* sm_tail = sm; 16 ListNode* bg = new ListNode(0); 17 ListNode* bg_tail = bg; 18 while(head){ 19 if(head->val < x){ 20 sm_tail->next = head; 21 //head = head->next; 22 sm_tail = sm_tail->next; 23 head = head->next; 24 sm_tail->next = NULL; 25 } 26 else{ 27 bg_tail->next = head; 28 head = head->next; 29 bg_tail = bg_tail->next; 30 //head = head->next; 31 bg_tail->next = NULL; 32 } 33 //head = head->next; 34 } 35 sm_tail->next = bg->next; 36 return sm->next; 37 } 38 };
标签：
原文地址：http://www.cnblogs.com/amazingzoe/p/4512677.html
