Sort List

最新推荐文章于 2025-08-19 16:19:29 发布

weixin_34037515

最新推荐文章于 2025-08-19 16:19:29 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/nnoth/p/3766036.html

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;
       deque<ListNode*> nodeVec;
       ListNode *root = new ListNode(-1);
       while(head){
           nodeVec.push_back(head);
           head = head->next;
       }
       buildHeap(nodeVec);
       int size = nodeVec.size()-1;
       head= extractMin(nodeVec,size);
       root->next = head;
     
       while(size>-1){
           head->next = extractMin(nodeVec,size);
           head = head->next;
       }
       head->next=NULL;
       head = root->next;
       delete root;
       return head;
    }
    ListNode* extractMin(deque<ListNode*> &nodeVec,int &size){
        ListNode *minNode = nodeVec[0];
        int index = 0;
        exchange(nodeVec,index,size);
        --size;
        MIN_Heapity(nodeVec,index,size);
        return minNode;
    }
    void buildHeap(deque<ListNode*> &nodeVec){
        int size = nodeVec.size()-1;
        for(int i = size/2; i>=0; --i){
            MIN_Heapity(nodeVec,i,size);
        }
    }
    void MIN_Heapity(deque<ListNode*> &nodeVec, int &p,int &size){
        if(size<=0)
            return;
        int min,index;
        index = p;
        min = nodeVec[p]->val;
        if(2*p+2<=size && min>nodeVec[2*p+2]->val)
        {
            min = nodeVec[2*p+2]->val;
            index = 2*p+2;
        }
        if(2*p+1<=size && min>nodeVec[2*p+1]->val){
			min = nodeVec[2*p+1]->val;
            index = 2*p+1;
        }
        if(index!=p){
            exchange(nodeVec,index,p);
            MIN_Heapity(nodeVec,index,size);
        }
    }
    void exchange(deque<ListNode*> &nodeVec, int &i,int &j){
        ListNode* tmp = nodeVec[i];
        nodeVec[i] = nodeVec[j];
        nodeVec[j] = tmp;
    }
};

　　TIPS:

用堆排序来做。

本来用快排来做，但最坏情况为O(N ^2)，即使采用随机快排，时间代价也不稳定，尤其对于有大量重复的例子，据说可以去重来处理

如果用归并排序，开全局数组N，我发现上面的写法也用了一个N vector的空间。。再改改

转载于:https://www.cnblogs.com/nnoth/p/3766036.html