No.148 Sort List

No.148 Sort List

Sort a linked list in O(n log n) time using constant space complexity.

分析：

　　常量空间且O(nlogn)时间复杂度，单链表适合用归并排序，双向链表适合用快速排序

　　有一个问题是：若算上栈空间，空间复杂度也为O(nogn)

　　还是对排序算法不够熟练！！

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x):val(x),next(NULL){}
};
class Solution
{
public:
    ListNode* sortList(ListNode* head)
    {//单链表排序：要求时间复杂度为O(nlogn),空间复杂度为O(1)
        if(head == NULL || head->next == NULL)//终止条件
            return head;

        //快慢指针找到中间节点
        ListNode *fast = head;
        ListNode *slow = head;
        //将链表分为两段，因为fast快，所以以它为判断条件
        while(fast->next != NULL && fast->next->next != NULL)
        {
            slow = slow->next;
            fast = fast->next->next;
        }

        ListNode *second = slow->next;
        slow->next = NULL;//断开链表
        //错：sortList(head);
        //错：sortList(second);
        //错：MergeSortedList(head, second);
        //错：return head;
        ListNode *l1= sortList(head);
        ListNode *l2 = sortList(second);
        return MergeSortedList(l1, l2);
    }

    ListNode* MergeSortedList(ListNode *first, ListNode *second)
    {
        if(first == NULL)
            return second;
        if(second == NULL)
            return first;

        ListNode *head, *current;
        if(first->val <= second->val)
        {
            head = first;
            first = first->next;
        }
        else
        {
            head = second;
            second = second->next;
        }
        current = head;

        while(first!= NULL && second != NULL)
        {
            if(first->val <= second->val)
            {
                current->next = first;
                first = first->next;
            }
            else
            {
                current->next = second;
                second = second->next;
            }
            current = current->next;
        }
/*
　　　　　　//注意与合并有序数组的区别
        while(first != NULL)
        {
            current->next = first;
            first = first->next;
            current = current->next;
        }
        while(second != NULL)
        {
            current->next = second;
            second = second->next;
            current = current->next;
        }
*/
        if(first)
            current->next = first;
        if(second)
            current->next = second;

        return head;
    }
};