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完善以下代码，以给定x为基准，将链表分割为两部分，所有小于x

[问答题]

完善以下代码，以给定x为基准，将链表分割为两部分，所有小于x的结点排在大于或等于x的结点之前。
public LinkedListNode partition(LinkedListNode node, int x)
{
  LinkedListNode beforeStart = null;
  LinkedListNode afterStart = null;

  while( node != null)
  {
LinkedListNode next =   1  ;
if (node.data < x)
{
  node.next =   2   ;
  beforeStart = node;
}
else{
     3  = afterStart;
  afterStart = node;
}
node = next
  }
  if(beforeStart == null){ return afterStart;}

  LinkedListNode head = beforeStart;
  while(beforeStart.next != null){
       4     = beforeStart.next;
  }
  beforeStart.next =   5   ;
  return head;
}

牛客379544753号

public LinkedListNode partition(LinkedListNode node, int x)
{
  LinkedListNode beforeStart = null;
  LinkedListNode afterStart = null;

  while( node != null)
  {
LinkedListNode next = node.next ;
if (node.data < x)
{
  node.next =   beforeStart ;
  beforeStart = node;
}
else{
     node.next = afterStart;
  afterStart = node;
}
node = next
  }
  if(beforeStart == null){ return afterStart;}
  LinkedListNode head = beforeStart;
  while(beforeStart.next != null){
       beforeStart = beforeStart.next;
  }
  beforeStart.next =   afterStart ;
  return head;
}

发表于 2022-02-18 10:47:05 回复(0)

知道自己很菜

public Node partition(int k) {
    Node bStart = null,bEnd = null,aStart = null,aEnd = null;  Node cur = this.head;   while (cur != null) { if (cur.data < k) { if (bStart == null) {
                bStart = cur;  bEnd = cur;  }else {
                bEnd.next = cur;  bEnd = bEnd.next;  }
        }else { if (aStart == null) {
                aStart = cur;  aEnd = cur;  }else {
                aEnd.next = cur;  aEnd = aEnd.next;  }
        }
        cur = cur.next; //自己写的时候忘了    没有这句cur无法往下走  } //拼接  //bStart = null，返回aStart   bStart != null 才需要拼接  //bEnd.next需要为空  if (bStart == null) { return aStart;  }else {
        bEnd.next = aStart;  aEnd.next = null;  } return bStart; }