1.问题描述
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.
来自 https://leetcode.com/problems/partition-list/description/
2.题目分析
要求把链表中把>=x的元素放到链表末尾,逐个比较,如果符合条件则从链表中将此元素删除,并将删除的元素用尾插法加入新的链表。直到遍历完链表,再将两个链表连接起来。
3.C++代码
//我的代码:(beats 87%)
ListNode* partition(ListNode* head, int x)
{
if (head == NULL)return head;
ListNode*dummy = new ListNode(0);
dummy->next = head;
ListNode*p = dummy;
ListNode*r = new ListNode(0);//新链表存储>=x的值
ListNode*tail = r;
while (p->next)
{
if (p->next->val >= x)
{
tail->next = p->next;
p->next = p->next->next;
tail = tail->next;
tail->next = NULL;
}
else
p = p->next;
}
p->next = r->next;//链表连接
return dummy->next;
}
//另一种方法:(beats 87%)
//分两个区间left和right;
ListNode* partition2(ListNode* head, int x)
{
if (head == NULL)return head;
ListNode*leftdummy = new ListNode(0);
ListNode*rightdummy = new ListNode(0);
ListNode*leftcur = leftdummy;
ListNode*rightcur = rightdummy;
while (head)
{
if (head->val < x)
{
leftcur->next = head;
leftcur = leftcur->next;
}
else
{
rightcur->next = head;
rightcur = rightcur->next;
}
head = head->next;
}
leftcur->next = rightdummy->next;