Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
想法很简单，就是两两合并。在Merge Two Sorted Lists这道题已经实现了两两合并的代码了，就直接拿过来用。
假设l[i]是第i条链的长度，那么共有k-1次合并，最坏情况下每次合并的时间复杂度为O(l[i - 1] + l[i])，0 < i < k，因此最坏情况下总的复杂度为O(l[0]+2*l[1]+...+2*l[k - 2] +l[k-1])。如果链的最大长度为n，那么算法复杂度就是O(2*n*(k-1))=O(nk)。
如果用二分法的话， 也就是将所有的链分成两份，每份各自合并完之后再合并起来。空间复杂度会是O(logk)， 时间复杂度T(k)=2T(k/2) +O(nk)，由主定理得O(nklgk)。
主定理见wiki。
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 if (lists.empty()) return NULL; 13 ListNode* head = lists[0]; 14 for (int i = 1; i < lists.size(); ++i) { 15 head = mergeTwoLists(head, lists[i]); 16 } 17 18 return head; 19 } 20 21 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { 22 ListNode *head, *t3; 23 head = t3 = new ListNode(0); 24 while (l1 != NULL && l2 != NULL) { 25 if (l1->val <= l2->val) { 26 t3->next = l1; 27 l1 = l1->next; 28 } else { 29 t3->next = l2; 30 l2 = l2->next; 31 } 32 t3 = t3->next; 33 } 34 35 while (l1 != NULL) { 36 t3->next = l1; 37 t3= t3->next; 38 l1 = l1->next; 39 } 40 while (l2 != NULL) { 41 t3->next = l2; 42 t3 = t3->next; 43 l2 = l2->next; 44 } 45 46 ListNode *ret = head->next; 47 delete head; 48 return ret; 49 } 50 };
Leetcode | Merge k Sorted Lists,布布扣,bubuko.com
Leetcode | Merge k Sorted Lists
原文：http://www.cnblogs.com/linyx/p/3704473.html