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.
Input: head = 1->4->3->2->5->2, x = 3 Output: 1->2->2->4->3->5
题意：
给定一个链表和一个值，把小于等于和大于该值的部分分别放到链表的一前一后去。
思路：
先分为两个链表
然后合并
代码：
1 class Solution { 2 public ListNode partition(ListNode head, int x) { 3 ListNode leftDummy = new ListNode(-1); 4 ListNode rightDummy = new ListNode (-1); 5 ListNode left_cur = leftDummy; 6 ListNode right_cur = rightDummy; 7 ListNode cur = head; 8 9 while( cur != null){ 10 if(cur.val < x){ 11 left_cur.next = cur; 12 left_cur = cur; 13 }else{ 14 right_cur.next = cur; 15 right_cur = cur; 16 } 17 cur = cur.next; 18 } 19 left_cur.next = rightDummy.next; 20 right_cur.next = null; 21 return leftDummy.next; 22 }
转载于:https://www.cnblogs.com/liuliu5151/p/9227167.html
