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.
Example
Given 1->4->3->2->5->2->null and x = 3,
return 1->2->2->4->3->5->null.
1 /** 2 * Definition for ListNode. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int val) { 7 * this.val = val; 8 * this.next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 /** 14 * @param head: The first node of linked list. 15 * @param x: an integer 16 * @return: a ListNode 17 */ 18 public ListNode partition(ListNode head, int x) { 19 ListNode left = new ListNode(0); 20 ListNode right = new ListNode(0); 21 22 ListNode leftHead = left; 23 ListNode rightHead = right; 24 25 while (head != null) { 26 if (head.val < x) { 27 left.next = head; 28 left = head; 29 } else { 30 right.next = head; 31 right = head; 32 } 33 head = head.next; 34 } 35 left.next = rightHead.next; 36 right.next = null; 37 return leftHead.next; 38 } 39 }
转载请注明出处：cnblogs.com/beiyeqingteng/
标签：
原文地址：http://www.cnblogs.com/beiyeqingteng/p/5636536.html
