Leetcode 148. Sort List (Medium) (cpp)

Niko_Ke

于 2016-08-17 09:39:16 发布

阅读量373

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/Niko_Ke/article/details/52227963

C++ 同时被 3 个专栏收录

299 篇文章

订阅专栏

Leetcode C++

287 篇文章

订阅专栏

Leetcode Linked List

26 篇文章

订阅专栏

本文介绍了一种使用常数空间复杂度，在O(nlogn)时间内对链表进行排序的方法。通过递归地将链表分为两部分，然后对每一部分进行排序，并最终合并两个有序链表。

Leetcode 148. Sort List (Medium) (cpp)

Tag: Linked List, Sort

Difficulty: Medium

/*

148. Sort List (Medium)

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

*/
/**
 * 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 *newhead = split(head);
        return sort_combine(sortList(head), sortList(newhead));
    }
private:
    ListNode* split(ListNode* head) {
        ListNode *s = head, *f = head;
        while(f -> next != NULL && f -> next -> next != NULL) {
            s = s -> next;
            f = f -> next -> next;
        }
        f = s -> next;
        s -> next = NULL;
        return f;
    }
    ListNode* sort_combine(ListNode* head, ListNode* s) {
        ListNode *head_new = new ListNode(-1), *p = head_new;
        while (head != NULL || s != NULL) {
            if (head == NULL || s != NULL && head -> val > s -> val) {
                p -> next = s;
                s = s -> next;
            } else {
                p -> next = head;
                head = head -> next;
            }
            p = p -> next;
        }
        return head_new -> next;
    }
};