题目描述:
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.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
代码：
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
if(head == nullptr)
return head;
ListNode dummy1 = ListNode(-1);
ListNode dummy2 = ListNode(1);
ListNode *last1 = &dummy1;
ListNode *last2 = &dummy2;
ListNode *p = head;
while(p!=nullptr)
{
ListNode *next = p->next;
p->next = nullptr;
if(p->val<x)
{
last1->next = p;
last1 = p;
}
else
{
last2->next = p;
last2 = p;
}
p = next;
}
last1->next = dummy2.next;
return dummy1.next;
}
};
原文：https://www.cnblogs.com/zjuhaohaoxuexi/p/11787685.html