题目链接:https://leetcode.com/problems/partition-list/
题意清晰,类似于快排中的partition操作,只是pivot是给定的,而不是链表中的元素。
思路一:遍历整个链表,用两个ArrayList分别存储小于x的节点和大于等于x的节点,存储的过程中改变节点的指向以形成两个链表,最后将这两个链表连在一起。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
if(head==null || head.next==null)
return head;
ArrayList less=new ArrayList<>();
ArrayList equalgreater=new ArrayList<>();
ListNode p=head;
while(p!=null)
{
if(p.val0)
less.get(n-1).next=p;
less.add(p);
}
else
{
int n=equalgreater.size();
if(n>0)
equalgreater.get(n-1).next=p;
equalgreater.add(p);
}
p=p.next;
}
if(less.size()==0)
return equalgreater.get(0);
else if(equalgreater.size()==0)
return less.get(0);
else
{
less.get(less.size()-1).next=equalgreater.get(0);
equalgreater.get(equalgreater.size()-1).next=null;
return less.get(0);
}
}
}
O(n)的时间和空间复杂度,效率很一般,beat 35%。
思路二:Solutions里的双指针解法跟思路一本质上是一致的,代码如下:
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode before_head = new ListNode(0);
ListNode before = before_head;
ListNode after_head = new ListNode(0);
ListNode after = after_head;
while (head != null) {
if (head.val < x) {
before.next = head;
before = before.next;
} else {
after.next = head;
after = after.next;
}
head = head.next;
}
after.next = null;
before.next = after_head.next;
return before_head.next;
}
}