86. Partition List

86. Partition List

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.

解析

• 思路：新建两个节点preHead1与preHead2，分别为指向两个链表的头结点。把节点值小于x的节点链接到链表1上，节点值大等于x的节点链接到链表2上。最后把两个链表相连即可
• right.next=null;//这句很重要！链表最后一个元素如果小于x的话，那么right.next不为null; 自己的没有想到，画个简单的示意图就知道，为甚末要置NULL，后面就绕开了这个问题，分离每一个节点

// 86. Partition List
class Solution_86 {
public:
	ListNode* partition(ListNode* head, int x) {

		if (!head||!head->next)
		{
			return head;
		}
	
		ListNode*cur = head;
		ListNode*left = new ListNode(0);
		ListNode*p = left;
		ListNode*right = new ListNode(0);
		ListNode*q = right;
		while (cur)
		{
			ListNode* temp = cur;	
			cur = cur->next;
			temp->next = NULL;

			if (temp->val<x)
			{
				left->next = temp;
				left = left->next;
			}
			else
			{
				right->next = temp;
				right = right->next;
			}
		}

		left->next = q->next;
		return p->next;
	}
};

题目来源

• 86. Partition List