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题目要求：
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
这道题目题意是要将两个有序的链表合并为一个有序链表。为了编程方便，在程序中引入dummy节点。具体程序如下：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { 12 ListNode *dummy = new ListNode(0); 13 ListNode *head = nullptr, *start = dummy; 14 int flag = true; 15 while(l1 && l2) 16 { 17 if(l1->val < l2->val) 18 { 19 start->next = l1; 20 l1 = l1->next; 21 } 22 else 23 { 24 start->next = l2; 25 l2 = l2->next; 26 } 27 start = start->next; 28 } 29 30 if(!l1) 31 start->next = l2; 32 else if(!l2) 33 start->next = l1; 34 35 head = dummy->next; 36 delete dummy; 37 dummy = nullptr; 38 39 return head; 40 } 41 };
题目要求：
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
这道题本来想用比较直观的方法，但超时了。具体程序如下：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* mergeKLists(vector<ListNode*>& lists) { 12 int sz = lists.size(); 13 if(sz == 0) 14 return nullptr; 15 16 ListNode *node = new ListNode(0); 17 ListNode *head = nullptr, *start = node; 18 while(true) 19 { 20 int count = 0; 21 int minVal = INT_MAX; 22 int minNode = -1; 23 for(int i = 0; i < sz; i++) 24 { 25 if(lists[i] && lists[i]->val < minVal) 26 { 27 minNode = i; 28 minVal = lists[i]->val; 29 } 30 else if(!lists[i]) 31 count++; 32 } 33 34 if(count != sz) 35 { 36 start->next = lists[minNode]; 37 lists[minNode] = lists[minNode]->next; 38 } 39 else 40 { 41 head = node->next; 42 delete node; 43 node = nullptr; 44 45 return head; 46 } 47 48 start = start->next; 49 } 50 } 51 };
后来参考网上的做法，用Divide and Conquer方法会比较好，具体程序如下：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *mergeTwoLists(ListNode* l1, ListNode* l2) { 12 ListNode *node = new ListNode(0); 13 ListNode *head = nullptr, *start = node; 14 int flag = true; 15 while(l1 && l2) 16 { 17 if(l1->val < l2->val) 18 { 19 start->next = l1; 20 l1 = l1->next; 21 } 22 else 23 { 24 start->next = l2; 25 l2 = l2->next; 26 } 27 start = start->next; 28 } 29 30 if(!l1) 31 start->next = l2; 32 else if(!l2) 33 start->next = l1; 34 35 head = node->next; 36 delete node; 37 node = nullptr; 38 39 return head; 40 } 41 42 ListNode *mergeKLists(vector<ListNode*>& lists, int low, int high) 43 { 44 if(low < high) 45 { 46 int mid = (low + high) / 2; 47 return mergeTwoLists(mergeKLists(lists, low, mid), mergeKLists(lists, mid + 1, high)); 48 } 49 return lists[low]; 50 } 51 52 ListNode* mergeKLists(vector<ListNode*>& lists) { 53 int sz = lists.size(); 54 if(sz == 0) 55 return nullptr; 56 57 return mergeKLists(lists, 0, sz - 1); 58 } 59 };
算法分析摘自同一博文：
我们来分析一下上述算法的时间复杂度。假设总共有k个list，每个list的最大长度是n，那么运行时间满足递推式T(k) = 2T(k/2)+O(n*k)。根据主定理，可以算出算法的总复杂度是O(nklogk)。如果不了解主定理的朋友，可以参见主定理-维基百科。空间复杂度的话是递归栈的大小O(logk)。
LeetCode之“链表”：Merge Two Sorted Lists && Merge k Sorted Lists
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原文地址：http://www.cnblogs.com/xiehongfeng100/p/4601922.html
