"Leetcode Problem Solving Report" single-linked list sorting problems

leetcode Merge Sorted Lists

Problem description

Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.

Analysis and Solution

There is not much explanation for this problem, direct merging; look at the code:

1 /**2 * Definition for singly-linked list.3 * struct ListNode4  * {5 * int val;6 * ListNode *next;7 * ListNode (int x): Val (x), Next (NULL) {}8  * };9  */Ten classSolution One { A  Public: -listnode* mergetwolists (listnode* L1, listnode*L2) -     { theListNode Dummy (-1);  -ListNode *p = &dummy; -          while(L1 &&L2) -         { +             if(L1->val < l2->val) -             { +P->next =L1; AL1 = l1->Next; at             } -             Else  -             { -P->next =L2; -L2 = l2->Next; -             } inp = p->Next; -         } toP->next = (L1! = NULL)?L1:l2; +         returnDummy.next; -     } the};

View Code Leetcode-Merge k Sorted Lists

Problem description

Merge K sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Analysis and Solution

According to the idea of merging. The reference code is as follows:

1 classSolution2 {3 Private:4listnode* mergetwolists (listnode* L1, listnode*L2)5     {6ListNode Dummy (-1); 7ListNode *p = &dummy;8          while(L1 &&L2)9         {Ten             if(L1->val < l2->val) One             { AP->next =L1; -L1 = l1->Next; -             } the             Else  -             { -P->next =L2; -L2 = l2->Next; +             } -p = p->Next; +         } AP->next = (L1! = NULL)?L1:l2; at         returnDummy.next; -     } -  Public: -listnode* mergeklists (vector<listnode*>&lists) -     { -         intn =lists.size (), I, J; in         if(n = =0)returnNULL; -          while(N >1) to         { +              for(i =0, j = N1; I < J; i++, j--) -             { theLists[i] =mergetwolists (Lists[i], lists[j]); *             } $n = (n+1) /2; Panax Notoginseng         } -         returnlists[0];  the     } +};

View Code leetcode 147 insertion Sort List

Problem description

Sort a linked list using insertion sort.

Analysis and Solution

To insert a sort, refer to the following code:

1 classSolution2 {3  Public:4listnode* Insertionsortlist (listnode*head)5     {6         if(!head | |!head->next)returnhead;7ListNode Dummy (-1); Dummy.next =head;8ListNode *p = &dummy, *q = head, *pe =head;9          while(q)Ten         { One             if(Q->val >= pe->val) A             { -PE =Q; -Q = q->Next; the                 Continue;  -             } -              for(p = &dummy; P->next! = q; p = p->next) -                 if(P->next->val > Q->val) Break; +Pe->next = q->Next; -Q->next = p->Next; +P->next =Q; AQ = pe->Next; at         } -         returnDummy.next; -     } -};

View Code Leetcode 148 Sort List

Problem description

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

Analysis and Solution

Merge sort ideas, recursive implementation, the reference code is as follows:

1 classSolution2 {3  Public:4listnode* Sortlist (listnode*head)5     {6         if(!head | |!head->next)returnhead;7ListNode *fast = head, *slow =head;8          while(Fast->next && fast->next->next)9         {TenFast = Fast->next->Next; Oneslow = slow->Next; A         } -Fast = Slow->Next; -Slow->next =NULL; the         returnMerge (Sortlist (head), sortlist (Fast)); -     } - Private: -listnode* merge (ListNode *l1, ListNode *L2) +     { -ListNode Dummy (0);  +ListNode *p = &dummy; A          while(L1! = NULL && L2! =NULL) at         { -             if(L1->val < l2->val) -             { -P->next =L1; -L1 = l1->Next; -             } in             Else  -             { toP->next =L2; +L2 = l2->Next; -             } thep = p->Next; *         } $P->next = (L1! = NULL)?L1:l2;Panax Notoginseng         returnDummy.next; -     } the};

View Code

"Leetcode Problem Solving Report" single-linked list sorting problems