Partition List
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
.
把小于x的节点连接成一个链表less大于等于x的节点连接成great链表然后链接在一起
代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* lesshead=new ListNode(0);
ListNode* greathead=new ListNode(1);
ListNode* less=lesshead;
ListNode* great=greathead;
ListNode* cur=head;
while(cur)
{
if ((cur->val)<x)
{
lesshead->next=cur;
cur=cur->next;
lesshead=lesshead->next;
lesshead->next=NULL;
}
else
{
greathead->next=cur;
cur=cur->next;
greathead=greathead->next;
greathead->next=NULL;
}
}
lesshead->next=great->next;
return less->next;
}
};