【Leetcode】86. Partition List

最新推荐文章于 2022-10-07 19:26:43 发布

原创最新推荐文章于 2022-10-07 19:26:43 发布 · 123 阅读

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本文介绍了一种链表分隔算法，该算法将链表中所有小于给定值x的节点置于大于或等于x的节点之前，同时保持各分区内的相对顺序不变。通过创建两个虚拟头节点，分别用于连接小于x的节点和大于等于x的节点，最终将两部分链接起来，形成有序的链表。

题目：

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:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

思路：
新建两个链表。l1和l2，比x小的都在l1
等于或者大于的都在l2，因为要保留原始位置吗
最后curr1.next = dummy2.next;curr2.next=null;

public ListNode partition(ListNode head, int x) {

’
ListNode dummy1 = new ListNode(0);
ListNode dummy2 = new ListNode(0);
ListNode curr1 = dummy1;
ListNode curr2 = dummy2;

while(head!=null){

   if(head.val<x){
       curr1.next = head;
       curr1 = head;
   }else{
       curr2.next = head;
       curr2 = head;
   }
   head = head.next;} curr2.next=null;   
 curr1.next = dummy2.next;
return dummy1.next;    
    
}