sort list--LeetCode

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

看到O(n log n)的排序算法，适合单链表的首先想到的就是归并排序

达到这个复杂度的排序算法，对于单链表来说，最合适的就是

 <pre name="code" class="cpp">#include <iostream>#include <vector>using namespace std;/*对于链表的排序  使用归并排序最合适 */typedef struct list_node List;struct list_node{struct list_node* next;int value;};void print_list(List* list){List* tmp=list;while(tmp != NULL){cout<<tmp->value<<endl;tmp = tmp->next; }}/*初始化List  将从1~n的数字插入到链表中 */void Init_List(List*& head,int* array,int n){head = NULL;List* tmp;List* record; for(int i=1;i<=n;i++){tmp = new List;tmp->next = NULL;tmp->value = array[i-1];if(head == NULL){head = tmp;record = head;}else{record->next = tmp;record = tmp;}}}int Len_list(List* list){if(list == NULL)return 0;elsereturn Len_list(list->next)+1;}/*对于平常的merge操作 采用的分治方法 先分开然后再合并 */void FindMid(List*& list,List*& pre,List*& last){pre  = list;last = list->next;while(last != NULL && last->next != NULL){pre = pre->next;last = last->next;if(last->next != NULL)last = last->next;}last = pre->next;pre->next  = NULL;pre = list;}void Merge(List*& list,List*& pre,List*& last){if(pre == NULL){list = last;return ;}if(last == NULL){list = pre;return ;}List* cur;List* tmp;if(pre->value > last->value)swap(pre,last);list = pre;cur = pre;while(cur->next != NULL && last != NULL){if(cur->next->value > last->value)  //插入元素{tmp = last->next;last->next = cur->next;cur->next = last;cur = last;last = tmp;}elsecur = cur->next; }if(last != NULL) cur->next = last;}void Merge_sec(List*& list,List*& pre,List*& last){List* tmp = new List;list = tmp;while(pre != NULL && last != NULL){if(pre->value < last->value){tmp->next = pre;pre = pre->next;} else{tmp->next = last;last = last->next;} tmp = tmp->next;}if(last != NULL)tmp->next = last;elsetmp->next = pre;list = list->next; } void MergeSort(List*& list){if(list ==NULL || list->next == NULL)return ;List* pre=NULL;List* last=NULL;FindMid(list,pre,last); //找到中间点 将一个链表list从中间分成pre和last两部分  MergeSort(pre); //归并排序前半部分  MergeSort(last); //归并排序后半部分 Merge(list,pre,last); //将两部分的链表进行合并 }int main() {int array[]={7,4,9,15,2,1,6,10,12,11};List* head;Init_List(head,array,sizeof(array)/sizeof(array[0]));MergeSort(head);print_list(head);return 0;}

思路比较清晰，使用伪指针的一个方法

public ListNode sortList(ListNode head) {    return mergeSort(head);}private ListNode mergeSort(ListNode head){    if(head == null || head.next == null)        return head;    ListNode walker = head;    ListNode runner = head;    while(runner.next!=null && runner.next.next!=null)    {        walker = walker.next;        runner = runner.next.next;    }    ListNode head2 = walker.next;    walker.next = null;    ListNode head1 = head;    head1 = mergeSort(head1);    head2 = mergeSort(head2);    return merge(head1, head2);}private ListNode merge(ListNode head1, ListNode head2){    ListNode helper = new ListNode(0);    helper.next = head1;    ListNode pre = helper;    while(head1!=null && head2!=null)    {        if(head1.val<head2.val)        {            head1 = head1.next;        }        else        {            ListNode next = head2.next;            head2.next = pre.next;            pre.next = head2;            head2 = next;        }        pre = pre.next;    }    if(head2!=null)    {        pre.next = head2;    }    return helper.next;}