[Leetcode] Merge K sorted lists

最新推荐文章于 2024-04-03 21:24:25 发布

liuwenyirachel

最新推荐文章于 2024-04-03 21:24:25 发布

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本文深入探讨了如何将多个已排序的链表合并为一个有序链表的过程，并分析了算法的时间复杂度。采用二分法策略优化效率，确保合并操作的一次性完成。

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

二分法可以降低复杂度。一遍通过，耶~

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(lists.size()==0) return NULL;
        int size = lists.size();
        while(size>1)
        {
            int halfSize = (1+size)/2;
            for(int i=0;i<halfSize && halfSize+i<size;i++) //重要！
            {
                ListNode *l1 = lists[i];
                ListNode *l2 = lists[i+halfSize];
                ListNode *result = mergeTwoLists(l1,l2);
                lists[i] = result;
            }
            size = halfSize;
            
        }
        return lists[0];
        
    }
    
    ListNode *mergeTwoLists(ListNode* list1,ListNode* list2)
    {
        ListNode* tmp1 = list1;
        ListNode* tmp2 = list2;
        ListNode* vHead = new ListNode(0);
        ListNode* tmp = vHead;
        
        while(tmp1&&tmp2)
        {
            if(tmp1->val<tmp2->val)
            {
                tmp->next = tmp1;
                tmp1 = tmp1->next;
            }
            else
            {
                tmp->next = tmp2;
                tmp2 = tmp2->next;
            }
            tmp = tmp->next;
        }        
        while(tmp1)
        {
            tmp->next = tmp1;
            tmp1 = tmp1->next;
            tmp = tmp->next;
        }
        while(tmp2)
        {
            tmp->next = tmp2;
            tmp2 = tmp2->next;
            tmp = tmp->next;
        }
        
        return vHead->next;
    }
};