Leetcode Sort List 解题报告

这道题用 mergesort，linkedlist的mergesort可以达到O(1) 的空间复杂度，而array就很难做到。mergesort是个递归算法，在每一步，需要把当前的linkedlist均分成两段，分别sort，然后merge。均分的过程容易出错。快指针(fast)需要先“走两步”，否则，如果linkedlist如果只有两个节点，那么将陷入死循环。

merge的过程中有的解法加入一个dummy node，这样merge的过程非常简单，只需要不断地扩展dummy node开头的链表即可。但是这样做的结果是空间复杂度不再是O(1)了。

#include <iostream>
using namespace std;

/**
* Definition for singly-linked list.
*/

struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

void printList(ListNode *head)
{
while (head)
{
cout<<head->val<<"\t";
head = head->next;
}
cout<<endl;
}

class Solution {
public:
ListNode* merge(ListNode *h1, ListNode *h2)
{
ListNode *head = h1;
ListNode *n1, *n2, *p1 = NULL;
while (h1 && h2)
{
n1 = h1->next;
n2 = h2->next;
if (h1->val < h2->val)
{
p1 = h1;
h1 = n1;
}
else
{
if (p1 != NULL)
{
p1->next = h2;
}
else
{
head = h2;
}
h2->next = h1;
p1 = h2;
h2 = n2;
}
}

if (p1 == NULL)
{
return h2;
}

while (h2)
{
p1->next = h2;
p1 = h2;
h2 = h2->next;
}

return head;
}

ListNode *sortList(ListNode *head)
{
// cout<<"debug:";
// printList(head);
if (head == NULL || head->next == NULL)
{
return head;
}

ListNode *slow = head, *fast = head->next->next;
while (fast != NULL)
{
slow = slow->next;
fast = fast->next;
if (fast)
{
fast = fast->next;
}
}
ListNode *p = slow->next;
slow->next = NULL;
head = sortList(head);
p = sortList(p);
return merge(head, p);
}
};

int main()
{
ListNode *head = new ListNode(1);
head->next = new ListNode(3);
head->next->next = new ListNode(5);
head->next->next->next = new ListNode(5);
cout<<"before: ";
printList(head);
Solution s;
head = s.sortList(head);
cout<<"after: ";
printList(head);
}