【LeetCode】023. Merge k Sorted Lists

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

题解：

　　归并思想。

Solution 1

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* mergeKLists(vector<ListNode*>& lists) {
         int n = lists.size();
         if (n < )
             return nullptr;
 
         for (int i = ; i < n ; i *= ) {
             int len = i;
             for (int j = ; j + len < n; j +=  *len) {
                 lists[j] = mergeTwoLists(lists[j], lists[j + len]);
             }
         }
 
         return lists[];
     }
     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
         ListNode dummy(-);
         ListNode *cur = &dummy;
         while (l1 && l2) {
             if (l1->val < l2->val) {
                 cur->next = l1;
                 l1 = l1->next;
             } else {
                 cur->next = l2;
                 l2 = l2->next;
             }
             cur = cur->next;
         }
         cur->next = l1 ? l1 : l2;
 
         return dummy.next;
     }
 };

Solution 2

　　利用最小堆，c++的优先队列priority_queue。

 class Solution {
 public:
     struct cmp {
         bool operator () (ListNode *a, ListNode *b) {
             return a->val > b->val;
         }
     };
     ListNode *mergeKLists(vector<ListNode *> &lists) {
         priority_queue<ListNode*, vector<ListNode*>, cmp> q;
         for (auto list : lists) {
             if (list)
                 q.push(list);
         }
 
         ListNode *head = NULL, *pre = NULL, *tmp = NULL;
         while (!q.empty()) {
             tmp = q.top();
             q.pop();
             if (!pre)
                 head = tmp;
             else
                 pre->next = tmp;
             pre = pre->next;
             // 此节点所在数组仍有元素，则添加进最小堆中
             if (tmp->next)
                 q.push(tmp->next);
         }
         return head;
     }
 };

转自：Grandyang