LeetCode - partition-list

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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,
Given1->4->3->2->5->2and x = 3,
return1->2->2->4->3->5.

题意：

给定一个链表和一个值x，对其进行分区，使所有小于x的节点都位于大于或等于x的节点之前。

您应该保留两个分区中每个节点的原始相对顺序。

解题思路:

1.新建两个结点temp1，temp2 ，指向两个链表的头结点

2.新建两个结点cur1，cur2,指向两个结点 temp1，temp2

3.将小于X结点值的结点放入链表1中

4.将大于等于x结点值的结点放入链表2中

5.最后将两个链表连接起来

public ListNode partition(ListNode head, int x) {
		 if(head == null) {
			 return null;
		 }
		 
		 ListNode temp1 = new ListNode(0);
		 ListNode cur1 = temp1;
		 ListNode temp2 = new ListNode(0);
		 ListNode cur2 = temp2;
		 
		 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 = temp2.next;
		 cur2.next = null;//链表2断开，最后结点指向null
		 return temp1.next;
	 }