标签：space next tor public else tno leetcode length node
一：解题思路
方法一：之前做过一道合并2个链表的题目，那么第一种方法就是将数组中的链表两两合并，得到最后的结果。Time:O(k*n),Space:O(1)
方法二：采用分治法，两两合拼。不断递归，最后只剩下一个链表。Time:O(n*log(k)),Space:O(log(k))
二：完整代码示例 (C++版和Java版)
方法一C++:
class Solution { private: ListNode* mergeTwoSortedLists(ListNode* l1, ListNode* l2) { ListNode* dummy = new ListNode(0); ListNode* p = dummy; while (l1 != NULL && l2 != NULL) { if (l1->val < l2->val) { p->next = l1; l1 = l1->next; } else { p->next = l2; l2 = l2->next; } p = p->next; } if (l1 != NULL) p->next = l1; if (l2 != NULL) p->next = l2; return dummy->next; } public: ListNode* mergeKLists(vector<ListNode*>& lists) { if (lists.size() == 0) return NULL; ListNode* result = NULL; for (ListNode* list : lists) { result = mergeTwoSortedLists(result,list); } return result; } };
方法一Java:
class Solution { private ListNode mergeTwoSortedLists(ListNode l1,ListNode l2) { ListNode dummy=new ListNode(0); ListNode p=dummy; while (l1!=null && l2!=null) { if(l1.val<l2.val) { p.next=l1; l1=l1.next; } else { p.next=l2; l2=l2.next; } p=p.next; } if(l1!=null) p.next=l1; if(l2!=null) p.next=l2; return dummy.next; } public ListNode mergeKLists(ListNode[] lists) { if(lists==null || lists.length==0) return null; ListNode result=null; for(ListNode list:lists) { result=mergeTwoSortedLists(result,list); } return result; } }
方法二C++:
class Solution { private: ListNode* mergeTwoSortedList(ListNode* l1, ListNode* l2) { ListNode* dummy = new ListNode(0); ListNode* p = dummy; while (l1 != NULL && l2 != NULL) { if (l1->val < l2->val) { p->next = l1; l1 = l1->next; } else { p->next = l2; l2 = l2->next; } p=p->next; } if (l1 != NULL) p->next = l1; if (l2 != NULL) p->next = l2; return dummy->next; } ListNode* merge(vector<ListNode*>& lists, int start, int end) { if (start == end) return lists[start]; if (start > end) return NULL; int mid = start + (end-start) / 2; ListNode* left = merge(lists,start,mid); ListNode* right = merge(lists,mid+1,end); return mergeTwoSortedList(left,right); } public: ListNode* mergeKLists(vector<ListNode*>& lists) { if (lists.size() == 0) return NULL; return merge(lists,0,lists.size()-1); } };
方法二Java:
class Solution { private ListNode mergeTwoSortedList(ListNode l1,ListNode l2) { ListNode dummy=new ListNode(0); ListNode p=dummy; while (l1!=null && l2!=null) { if(l1.val<l2.val) { p.next=l1; l1=l1.next; } else { p.next=l2; l2=l2.next; } p=p.next; } if(l1!=null) p.next=l1; if(l2!=null) p.next=l2; return dummy.next; } private ListNode merge(ListNode[] lists,int start,int end) { if(start==end) return lists[start]; if(start>end) return null; int mid=start+(end-start)/2; ListNode left=merge(lists,start,mid); ListNode right=merge(lists,mid+1,end); return mergeTwoSortedList(left,right); } public ListNode mergeKLists(ListNode[] lists) { if(lists==null || lists.length==0) return null; return merge(lists,0,lists.length-1); } }
标签：space next tor public else tno leetcode length node
原文地址：https://www.cnblogs.com/repinkply/p/12694187.html
