标签：des style class blog code ext
Description:
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：这道题目就是将链表中小于目标值的元素移到前面，链表中所有元素的相对位置都应该保持不变，其实就是一些
最基本的链表操作，但是这里有一个小trick, 因为每次都要往前看一个元素，以确定操作，所以在链表前面增加一个额外元素，
将所有的操作在循环内就搞定了！
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 == NULL) return head; 13 14 ListNode *newh = new ListNode(0); 15 newh->next = head; 16 17 ListNode *pass= newh, *insertpos=newh, *findnod; 18 while(pass->next!=NULL) 19 { 20 if(pass->next->val>=x) 21 pass = pass->next; 22 else if(insertpos->next == pass->next) 23 { 24 pass = pass->next; 25 insertpos = insertpos->next; 26 } 27 else{ 28 findnod = pass->next; 29 pass->next = findnod->next; 30 //pass = pass->next; 31 32 findnod->next = insertpos->next; 33 insertpos->next = findnod; 34 insertpos = insertpos->next; 35 } 36 } 37 38 return newh->next; 39 } 40 };
Leetcode:Partition List,布布扣,bubuko.com
标签：des style class blog code ext
原文地址：http://www.cnblogs.com/soyscut/p/3787584.html
