[leetcode 147]Insertion Sort List

最新推荐文章于 2022-04-09 13:42:00 发布

转载最新推荐文章于 2022-04-09 13:42:00 发布 · 311 阅读

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#leetcode #链表 #插入排序

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本文介绍了一种使用插入排序算法对单链表进行排序的方法，详细解释了算法实现过程并提供了代码示例。

Sort a linked list using insertion sort.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *insertionSortList(ListNode *head) {
        if(NULL == head || NULL == head->next){return head;}
        ListNode *runner = head;
        ListNode *prerunner = head;
        ListNode *handle = head->next;
        ListNode *prehandle = head;
        while(handle)
        {
            runner = head;
            prerunner = head;
            while(runner!= handle && handle->val>=runner->val)
            {
                prerunner = runner;
                runner = runner->next;  
            }
            if(runner == handle) 
            {
                prehandle = handle;handle = handle->next;
            }
            else 
            {
                prehandle->next = handle->next;
                if(runner == head)
                {
                    head = handle;
                    handle->next = runner;
                }
                else
                {
                    handle->next = runner;
                    prerunner->next = handle;
                }
                handle = prehandle->next;
            }
        }
        return head;
    }
};

class Solution {
public:
    ListNode *insertionSortList(ListNode *head) {
        if(NULL == head || NULL == head->next){return head;}
        ListNode firstnode(0);
        firstnode.next = head;
        ListNode *handle = head->next;
        ListNode *prehandle = head;
        while(handle)
        {
            ListNode *prerunner = &firstnode;
            ListNode *runner = firstnode.next; //head may change
            while(runner!= handle && handle->val>=runner->val)
            {
                prerunner = runner;
                runner = runner->next;  
            }
            if(runner == handle) 
            {
                prehandle = handle;handle = handle->next;
            }
            else 
            {
                prehandle->next = handle->next;
                handle->next = runner;
                prerunner->next = handle;
                handle = prehandle->next;
            }
        }
        return firstnode.next;
    }
};