Sort a linked list in O(n log n) time using constant space complexity.
逻辑正确，尾指针没有置空，调了好久……
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/** * 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)return
head; ListNode *temp = head; int
n =0; while(temp!=NULL) { n++; temp = temp->next; } head =merge(head,n); return
head; } ListNode *merge(ListNode *head, int
n) { if(n>2) { ListNode * temp =head; ListNode * first; ListNode * second; ListNode * tail = NULL; int
mid = (1+n)/2; for(int
i = 0; i < mid ; i++)temp =temp->next; first = merge(head,mid); second = merge(temp,n-mid); //for(int i = mid; i < n ; i++)temp =temp->next; tail = temp->next; ListNode *ft =first; ListNode *st =second; ListNode *me = new
ListNode(0); temp =me; while(ft!= NULL && st!= NULL) { if(ft->val < st->val) { temp->next =ft; temp = temp->next; ft=ft->next; temp->next =NULL; } else { temp->next =st; temp = temp->next; st=st->next; temp->next =NULL; } } if(ft!=NULL) { temp->next = ft; //while(ft->next!=NULL)ft =ft->next; //ft->next = tail; } else
if(st!=NULL) { temp->next = st; //while(st->next!=NULL)st =st->next; //st->next = tail; } return
me->next; } else
if(n == 2) { if(head->val > head->next->val ) { ListNode * temp =head; ListNode * second = head->next; temp->next = NULL; head = second; //temp->next = second->next; head->next =temp; return
head; } else { head->next->next=NULL; return
head; } } else
if(n == 1) { head->next =NULL; return
head; } } ListNode * sortl(ListNode *head , int
n) { if(n<2)return
head; ListNode * temp; ListNode * left =new
ListNode(0); ListNode * lt = left; ListNode * right = new
ListNode(0); ListNode * rt = right; ListNode * mid; mid = head; int
va = mid->val; int
ln = 0, rn = 0; temp = head->next; for(int
i =1 ; i < n-1 ; i++) { if(temp->val <= va) { lt->next = temp; lt =lt->next; temp =temp->next; ln++; } else { rt->next = temp; rt = rt->next; temp =temp->next; rn++; } } ListNode * tail = temp->next; if(temp->val <= va) { lt->next = temp; lt =lt->next; ln++; } else { rt->next = temp; rt = rt->next; rn++; } if(ln >0) { lt->next = mid; head = sortl(left->next,ln); } else head = mid; if(rn>0) { rt->next =tail; mid ->next= sortl(right->next,rn); } else mid->next=tail; return
head; }}; |
原文：http://www.cnblogs.com/pengyu2003/p/3621962.html