leetcode-Partition List
2014-07-02 10:50
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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
return
思路:三个指针记录位置:p:插入位置的前一个结点,pre: q的前一个结点;q:遍历结点
当q->val小于x时,插入,插入头结点前面时需进行特殊处理
否则,向前遍历。
代码:
ListNode *partition(ListNode *head, int x) {
ListNode *p=NULL;
ListNode *q=head;
ListNode *pre=NULL;
while(q != NULL)
{
if(q->val < x)
{
if(p==NULL)
{
if(head->val>=x)
{
pre->next=q->next;
q->next=head;
head=q;
q=pre->next;
p=head;
}
else
{
pre=q;
q=q->next;
}
}
else
{
pre->next=q->next;
q->next=p->next;
p->next=q;
q=pre->next;
p=p->next;
}
}
else
{
if(pre!=NULL && pre->val<x)
{
p=pre;
}
pre=q;
q=q->next;
}
}
return head;
}
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given
1->4->3->2->5->2and x = 3,
return
1->2->2->4->3->5.
思路:三个指针记录位置:p:插入位置的前一个结点,pre: q的前一个结点;q:遍历结点
当q->val小于x时,插入,插入头结点前面时需进行特殊处理
否则,向前遍历。
代码:
ListNode *partition(ListNode *head, int x) {
ListNode *p=NULL;
ListNode *q=head;
ListNode *pre=NULL;
while(q != NULL)
{
if(q->val < x)
{
if(p==NULL)
{
if(head->val>=x)
{
pre->next=q->next;
q->next=head;
head=q;
q=pre->next;
p=head;
}
else
{
pre=q;
q=q->next;
}
}
else
{
pre->next=q->next;
q->next=p->next;
p->next=q;
q=pre->next;
p=p->next;
}
}
else
{
if(pre!=NULL && pre->val<x)
{
p=pre;
}
pre=q;
q=q->next;
}
}
return head;
}
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