cherrychenlee

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原文地址:https://www.jianshu.com/p/4f3bf22ea9de

时间限制:1秒 空间限制:32768K

题目描述

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

我的代码

/**
 * 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==nullptr || head->next==nullptr)
            return head;
        ListNode* fast=head->next;
        ListNode* slow=head;
        while(fast&&fast->next){
            fast=fast->next->next;
            slow=slow->next;
        }
        ListNode* left=sortList(slow->next);
        slow->next=nullptr;
        ListNode* right=sortList(head);
        return merge(left,right);
    }
    ListNode* merge(ListNode* left,ListNode* right){
        ListNode tmp(0);
        ListNode* p=&tmp;
        while(left&&right){
            if(left->val<right->val){
                p->next=left;
                left=left->next;
            }
            else{
                p->next=right;
                right=right->next;
            }
            p=p->next;
        }
        if(left)
            p->next=left;
        if(right)
            p->next=right;
        return tmp.next;
    }//时复:O(nlogn),空复:O(logn)
};

运行时间:13ms
占用内存:1232k

/**
 * 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==nullptr || head->next==nullptr)
            return head;
        int len=lenofList(head);
        ListNode tmp(0);
        tmp.next=head;
        for(int i=1;i<len;i<<=1){
            ListNode* cur=tmp.next;
            ListNode* tail=&tmp;
            while(cur){
                ListNode* left=cur;
                ListNode* right=cut(left,i);
                cur=cut(right,i);
                tail->next=merge(left,right);
                while(tail->next)
                    tail=tail->next;
            }
        }
        return tmp.next;
    }
    int lenofList(ListNode* p){
        int len=0;
        while(p){
            len++;
            p=p->next;
        }
        return len;
    }
    ListNode* cut(ListNode* p, int len){
        while(p&&(--len)){
            p=p->next;
        }
        if(!p)
            return nullptr;
        ListNode* next=p->next;
        p->next=nullptr;
        return next;
    }
    ListNode* merge(ListNode* left,ListNode* right){
        ListNode tmp(0);
        ListNode* p=&tmp;
        while(left&&right){
            if(left->val<right->val){
                p->next=left;
                left=left->next;
            }
            else{
                p->next=right;
                right=right->next;
            }
            p=p->next;
        }
        if(left)
            p->next=left;
        if(right)
            p->next=right;
        return tmp.next;
    }//时复:O(nlogn),空复:O(1)
};

运行时间:12ms
占用内存:1004k

posted on 2019-05-08 14:07  cherrychenlee  阅读(76)  评论(0编辑  收藏  举报