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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.
这道题开始没什么思路，后面看到提示说用双指针，有点懂了
思路
找出第一个大于等于x的位置front和前一个位置behind,behind后移，如果behind.val < x在front后面插入值为x的节点
1 public class Solution { 2 public ListNode partition(ListNode head, int x) { 3 ListNode headNode = new ListNode(0); 4 headNode.next = head; 5 6 ListNode front = head; 7 ListNode behind = headNode; //双指针 8 9 while(front != null){ //找出第一个大于等于x的位置和前一个位置 10 if(front.val >= x) 11 break; 12 front = front.next; 13 behind = behind.next; 14 }//while 15 if(front == null) 16 return head; //如果原列表符合规定直接返回头结点 17 while(front != null &&front.next != null){ //遇到比x小的插入到behind后面，同时删除后面的节点 18 if(front.next.val < x){ 19 ListNode temp = new ListNode(front.next.val); 20 temp.next = behind.next; 21 behind.next = temp; 22 behind = temp; 23 front.next = front.next.next; //删除节点 24 continue; 25 }//if 26 front = front.next; 27 }//while 28 return headNode.next; 29 } 30 }
这是我在leetcode上面的第100题
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原文地址：http://www.cnblogs.com/luckygxf/p/4209428.html
