今天上午在K一道单链表排序题（目前LeetCode第四题），利用快速排序实现的版本过不了时间限制，不排除我实现中优化做得不够。Partition的算法和我在这篇blogC++实现的快速排序中使用的算法是类似的，都是参考自算法导论。虽然提交LeetCode的题目超时了，但是这个代码可以留作参考。至于题目，一会儿我用归并试试。
#include<iostream>
#include<ctime>
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
ListNode *sortList(ListNode *head) {
if (head == NULL)
return head;
ListNode* tail=head;
ListNode* end = tail;
while (tail->next != NULL) {
//if (tail->next->next == NULL)
// end = tail;
tail = tail->next;
}
qSortList(head, tail);
return head;
}
ListNode* Partition(ListNode* head, ListNode* tail)
{
ListNode* newHead = new ListNode(0);
newHead->next = head;
ListNode* ipt = newHead, *jpt = head;
int ar = tail->val;
while (jpt != tail)
{
if (jpt->val < ar)
{
ipt = ipt->next;
swap(ipt->val, jpt->val);
}
jpt = jpt->next;
}
ipt = ipt->next;
swap(ipt->val, tail->val);
return ipt;
}
void qSortList(ListNode*head, ListNode*tail)
{
if (head == NULL)
return;
if (tail == NULL)
return;
if (tail->next!=NULL && tail->next == head)
return;
else if (tail->next!=NULL &&tail->next->next!=NULL && tail->next->next== head)
return;
if (head == tail)
return;
if (head->next == tail)
{
if (head->val > tail->val)
swap(head->val,tail->val);
return;
}
ListNode* mid = Partition(head, tail);
ListNode*tail1 = head;
if (tail1!=mid)
while (tail1->next != mid) tail1 = tail1->next;
ListNode*head2 = mid->next;
qSortList(head, tail1);
qSortList(head2, tail);
}
};
int main()
{
ListNode* head0 = new ListNode(200);
ListNode* head = head0;
srand(time(NULL));
for (int i = 0; i < 10000; i++)
{
ListNode* mNode = new ListNode(rand() % 4);
head0->next = mNode;
head0 = head0->next;
}
Solution sln;
ListNode*res=sln.sortList(head);
while (res->next != NULL)
{
cout << res->val << " ";
res = res->next;
}
return 0;
}