[LeetCode 解題報告]023. Merge k Sorted Lists

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

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

考察點：在021. Merge Two Sorted Lists的基礎上，考察如何合併K個。只需兩兩合併即可。

/**
 * 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) {
        if (lists.empty())
            return NULL;
        int n = lists.size();
        while (n > 1) {
            int k = (n+1) / 2;
            for (int i = 0; i < n/2; i ++)
                lists[i] = mergeTwoLists(lists[i], lists[i+k]);
            n = k;
        }
        return lists[0];
        
    }
    
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode *dummy = new ListNode(-1), *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;
    }
};

 完，