Partition List leetcode java
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.
For example,
Given 1->4->3->2->5->2
and x = 3,
return 1->2->2->4->3->5
.
题解:
这道题就是说给定一个x的值,小于x都放在大于等于x的前面,并且不改变链表之间node原始的相对位置。每次看这道题我老是绕晕,纠结为什么4在3的前面。。其实还是得理解题意,4->3->5都是大于等3的数,而且这保持了他们原来的相对位置 。
所以,这道题是不需要任何排序操作的,题解方法很巧妙。
new两个新链表,一个用来创建所有大于等于x的链表,一个用来创建所有小于x的链表。
遍历整个链表时,当当前node的val小于x时,接在小链表上,反之,接在大链表上。这样就保证了相对顺序没有改变,而仅仅对链表做了与x的比较判断。
最后,把小链表接在大链表上,别忘了把大链表的结尾赋成null。
代码如下:
1 public ListNode partition(ListNode head, int x) {
2 if(head==null||head.next==null)
3 return head;
4
5 ListNode small = new ListNode(-1);
6 ListNode newsmallhead = small;
7 ListNode big = new ListNode(-1);
8 ListNode newbighead = big;
9
10 while(head!=null){
11 if(head.val<x){
12 small.next = head;
13 small = small.next;
14 }else{
15 big.next = head;
16 big = big.next;
17 }
18 head = head.next;
19 }
20 big.next = null;
21
22 small.next = newbighead.next;
23
24 return newsmallhead.next;
25 }
2 if(head==null||head.next==null)
3 return head;
4
5 ListNode small = new ListNode(-1);
6 ListNode newsmallhead = small;
7 ListNode big = new ListNode(-1);
8 ListNode newbighead = big;
9
10 while(head!=null){
11 if(head.val<x){
12 small.next = head;
13 small = small.next;
14 }else{
15 big.next = head;
16 big = big.next;
17 }
18 head = head.next;
19 }
20 big.next = null;
21
22 small.next = newbighead.next;
23
24 return newsmallhead.next;
25 }