[LeetCode] 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

.

类似于qsort的partition的思想，用两根指针记录左右两边。

/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (head == NULL)
return NULL;

ListNode *pPre = NULL;
ListNode *p = head;
ListNode *q = head;

while(q->next)
q = q->next;

ListNode *tail = q;

while(p != q)
{
if (p->val >= x)
{
ListNode *pNext = p->next;
if (head == p)
head = pNext;
tail->next = p;
p->next = NULL;
tail = p;
if (pPre)
pPre->next = pNext;
p = pNext;
}
else
{
pPre = p;
p = p->next;
}
}

if (p->val >= x && q != tail)
{
ListNode *pNext = p->next;
if (head == p)
head = pNext;
tail->next = p;
p->next = NULL;
if (pPre)
pPre->next = pNext;
}

return head;
}
};