86. Partition List

JIAKAOBO

LeetCode

感谢赞助!

• ㊗️
• 大家
• offer
• 多多！

Problem

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

Code

/**
 * 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) {
        ListNode dummy1 = new ListNode(0);
        ListNode curr1 = dummy1;
        ListNode dummy2 = new ListNode(0);
        ListNode curr2 = dummy2;

        while(head != null){
            ListNode temp = new ListNode(head.val);
            if(head.val < x){
                curr1.next = temp;
                curr1 = curr1.next;
            } else {
                curr2.next = temp;
                curr2 = curr2.next;
            }

            head = head.next;
        }
        curr1.next = dummy2.next;
        return dummy1.next;
    }
}