[LeetCode]86. Sort List链表排序

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

 

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解法：这题和Insertion Sort List链表插入排序一样，都是要求对单链表进行排序，但是本题要求时间复杂度在O(nlogn)，空间复杂度在O(1)。列出常用排序算法：冒泡排序、插入排序、选择排序、快速排序、堆排序、归并排序、基数排序、希尔排序等等，满足时间空间复杂度的有堆排序。但是对于链表，因为其不能随机存取，因此建堆时间复杂度可能会比较高。满足时间复杂度的还有快速排序和归并排序，快速排序中划分过程要求指针从两头往中间移动，这在链表中也不好实现。而归并排序的O(n)空间消耗在辅助数组，对于链表来说不需要这个辅助空间，因此可以用也只能用归并排序。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (head == NULL || head->next == NULL) return head;
        ListNode *slow = head, *fast = head;
        while (fast->next != NULL && fast->next->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode *head1 = head, *head2 = slow->next;
        slow->next = NULL; // 注意将链表分开为两部分
        head1 = sortList(head1);
        head2 = sortList(head2);
        head = mergeTwoLists(head1, head2);
        return head;
    }
private:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (l1 == NULL) return l2;
        if (l2 == NULL) return l1;
        ListNode* help = new ListNode(0);
        ListNode* head = help;
        while (l1 != NULL && l2 != NULL) {
            if (l1->val <= l2->val) {
                help->next = l1;
                l1 = l1->next;
            }
            else {
                help->next = l2;
                l2 = l2->next;
            }
            help = help->next;
        }
        if (l1 != NULL) help->next = l1;
        if (l2 != NULL) help->next = l2;
        return head->next;
    }
};