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.
the key point of this problem is "DO
NOT FORGET TO SET NULL".
?
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/**
* 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) {
if(head == NULL) return
NULL;
bool
bBig=false;
bool
bSmall=false;
ListNode *p=head;
ListNode * pBig = NULL;
ListNode * pSmall = NULL;
ListNode *pHead = NULL;
ListNode *pHead2 = NULL;
while(p!=NULL)
{
int
v = p->val;
if(v >= x)
{
if(pBig==NULL)
{
pBig = p;
pHead2 = p;
}
else
{
pBig->next=p;
pBig = pBig->next;
}
}
if(v < x)
{
if(pSmall==NULL)
{
pSmall = p;
pHead = p;
}
else
{
pSmall->next=p;
pSmall = pSmall->next;
}
}
p=p->next;
}
!!!!!!!!//this is really import, or there will be a circle in linklist!!!!!!!!!!
if(pSmall != NULL)
{
pSmall->next = NULL;
}
if(pBig != NULL)
{
pBig->next = NULL;
}
if(pHead != NULL)
{
pSmall->next = pHead2;
}
else
{
pHead = pHead2;
}
return
pHead;
}
};
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[LeetCode] [Partition List 2012-04-30]
时间： 2024-08-26 18:38:18
