LeetCode每天刷day17:Merge k Sorted Lists-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_36428171/article/details/88954681

本文介绍了一种使用分治思想合并多个已排序链表的方法，并提供了C++实现代码。该算法通过递归将问题规模减半，最终将两个排序链表合并为一个。适用于需要高效合并多个排序链表的场景。

题目:
合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* res = new ListNode(0);
        ListNode* head = res;
        while(l1!=NULL || l2!=NULL){
            if(l1&&l2){
                if(l1->val < l2->val){
                    head->next = l1;
                    l1 = l1->next;
                }
                else{
                    head->next = l2;
                    l2 = l2->next;
                }
                head = head->next;
            }
            while(l1!=NULL && l2==NULL){
                head->next = l1;
                head = head->next;
                l1 = l1->next;
            }
            while(l1==NULL && l2!=NULL){
                head->next = l2;
                head = head->next;
                l2 = l2->next;
            }
        }
        return res->next;
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(lists.empty()){
            return NULL;
        }
        if(lists.size() == 1){
            return lists[0];
        }
        else if(lists.size() == 2){
            return mergeTwoLists(lists[0], lists[1]);
        }
        int mid = lists.size()/2;
        vector<ListNode*> l1,l2;
        for(int i = 0; i < mid; i++)
            l1.push_back(lists[i]);
        for(int i = mid; i < lists.size(); i++)
            l2.push_back(lists[i]);
        ListNode *L1 = mergeKLists(l1);
        ListNode *L2 = mergeKLists(l2);
        return mergeTwoLists(L1, L2); 
    }
};

合并排序,分治思想运用