[LeetCode]Sort List

题目描述：

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

解题方案：

题目要求的时间复杂度是 O(n log n)，常数级空间复杂度。所以这里用了归并排序，归并排序在数组上操作比较方便，但是这里要排序的是链表。我们用到两个指针将链表一份为二，一个指针q每次前进两步，一个指针p每次前进一步，当q到达链表结尾时，p到达链表中间位置。下面是该题的代码：

class Solution {
public:
    ListNode *sortList(ListNode *head) {
        return mergeSort(head);
    }
    ListNode *mergeSort(ListNode *head) {
        if (!head || head->next == NULL) {
            return head;
        }
        ListNode *p = head;
        ListNode *q = head;
        ListNode *pre = NULL;
        while (q && q->next != NULL) {
            q = q->next->next;
            pre = p;
            p = p->next;
        }
        pre->next = NULL;
        ListNode *llist = mergeSort(head);
        ListNode *rlist = mergeSort(p);
        return merge(llist, rlist);
    }
    ListNode *merge(ListNode *lhs, ListNode *rhs) {
        ListNode *newhead = new ListNode(0);
        ListNode *p = newhead;
        while (lhs && rhs) {
            if (lhs->val <= rhs->val) {
                p->next = lhs;
                lhs = lhs->next;
            } else {
                p->next = rhs;
                rhs = rhs->next;
            }
            p = p->next;
        }
        if (lhs == NULL) {
            p->next = rhs;
        } else {
            p->next = lhs;
        }
        p = newhead->next ;
        newhead->next = NULL;
        delete newhead;
        return p;
    }
};