首先,因为最近要期末考,所以更新将会变的极为不规律,见谅(下次更新可能要到下周三或周四了)
这题首先是说给了一个串,和一个X
现在让你把它划分成两部分,所有小于x的放在左边,大于等于的x放在右边,形成一个新的链表
限制就是,原有的顺序不能动,就是比x小的相对的顺序,大于等于x的相对顺序也不能变
而这道题的做法其实也很简单,保持两个链表就可以了。
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.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */
public class Solution {
/** * 保存两个链就可以*/
public ListNode partition(ListNode head, int x) {
ListNode L = new ListNode(-1);
ListNode ME = new ListNode(-1);
ListNode LH=L,MEH=ME;
while(head!=null){
if (head.val < x){
L.next=head;
L=L.next;
} else{
ME.next=head;
ME=ME.next;
}
head=head.next;
}
L.next=MEH.next;
ME.next=null;
return LH.next;
}
}