Leetcode 23. Merge k Sorted Lists
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
题目意思：k个有序链表合并后仍为一个有序链表
解题思路：我先将链表存储在vector中，然后我用新的vector去存储链表中的每个节点，之后快速排序（stl std::sort）,然后再对排序过后的节点组成链表，不是最优解，但是一个比较容易想的方法，时间复杂度是O(kn*logkn)
1 #include <stdio.h> 2 3 struct ListNode { 4 int val; 5 ListNode *next; 6 ListNode(int x) : val(x), next(NULL) {} 7 }; 8 9 #include <vector> 10 #include <algorithm> 11 12 bool cmp(const ListNode *a, const ListNode *b){ 13 return a->val < b->val; 14 } 15 16 class Solution { 17 public: 18 ListNode* mergeKLists(std::vector<ListNode*>& lists) { 19 std::vector<ListNode *> node_vec; 20 for (int i = 0; i < lists.size(); i++){ 21 ListNode *head = lists[i]; 22 while(head){ 23 node_vec.push_back(head); 24 head = head->next; 25 } 26 } 27 if (node_vec.size() == 0){ 28 return NULL; 29 } 30 std::sort(node_vec.begin(), node_vec.end(), cmp); 31 for (int i = 1; i < node_vec.size(); i++){ 32 node_vec[i-1]->next = node_vec[i]; 33 } 34 node_vec[node_vec.size()-1]->next = NULL; 35 return node_vec[0]; 36 } 37 }; 38 39 int main(){ 40 ListNode a(1); 41 ListNode b(4); 42 ListNode c(6); 43 ListNode d(0); 44 ListNode e(5); 45 ListNode f(7); 46 ListNode g(2); 47 ListNode h(3); 48 a.next = &b; 49 b.next = &c; 50 d.next = &e; 51 e.next = &f; 52 g.next = &h; 53 Solution solve; 54 std::vector<ListNode *> lists; 55 lists.push_back(&a); 56 lists.push_back(&d); 57 lists.push_back(&g); 58 ListNode *head = solve.mergeKLists(lists); 59 while(head){ 60 printf("%d\n", head->val); 61 head = head->next; 62 } 63 return 0; 64 }
通过~
