[leetcode]86. Partition List@Java解題報告__Java

https://leetcode.com/problems/partition-list/description/

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.

題意：給定一個單鏈表和一個x，把鏈表中小於x的放到前面，大於等於x的放到後面，每部分元素的原始相對位置不變。

思路：建立兩個節點preHead1與preHead2，分別為指向兩個鏈表的頭結點。

把節點值小於x的節點連結到鏈表1上，節點值大等於x的節點連結到鏈表2上。

最後把兩個鏈表相連即可

package go.jacob.day817;/** * 86. Partition List *  * @author Jacob * */public class Demo3 {public ListNode partition(ListNode head, int x) {//preHead1與preHead2分別為兩個鏈表頭結點的前移節點（一個鏈表的節點值小於x，另一個大於等於x）ListNode preHead1 = new ListNode(0), preHead2 = new ListNode(0);ListNode cur1 = preHead1, cur2 = preHead2;while (head != null) {if (head.val < x) {cur1.next = head;cur1 = cur1.next;} else {cur2.next = head;cur2 = cur2.next;}head = head.next;}cur1.next = preHead2.next;cur2.next = null;return preHead1.next;}}