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目录

题目描述

方法一、k-1次两两合并

方法二、分治法合并

方法三、使用优先队列

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题目描述

23. Merge k Sorted Lists

方法一、k-1次两两合并

选第一个链表作为结果链表，每次将后面未合并的链表合并到结果链表中，经过k-1次合并，即可得到答案。假设每个链表的最长长度是n，时间复杂度O（n+2n+3n+...(k-1)n） = O(n) = O(n)。空间复杂度O(1)。

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* mergeKLists(vector<ListNode*>& lists) {int n = lists.size();if(n == 0)return nullptr;ListNode* ans = lists[0];for(int i = 1;i< n;i++){ans = merge(ans,lists[i]);}return ans;}ListNode* merge(ListNode* L1,ListNode* L2){ListNode* dummy = new ListNode();ListNode* cur = dummy;while(L1&&L2){if(L1->val < L2->val){cur->next = L1;cur = L1;L1 = L1->next;}else{cur->next = L2;cur = L2;L2 = L2->next;}}cur->next = L1 != nullptr ? L1 : L2;ListNode* res = dummy->next;delete dummy;return res;}
};

方法二、分治法合并

时间复杂度 O(kn×logk)。空间复杂度 O(logk) 。

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* mergeKLists(vector<ListNode*>& lists) {int n = lists.size();if(n == 0)return nullptr;return merge(lists,0,n-1);}ListNode* merge(vector<ListNode*>& lists,int left,int right){if(left == right)return lists[left];if(left>right)return nullptr;int mid = left + ((right-left)>>1);return mergeTwoList(merge(lists,left,mid),merge(lists,mid+1,right));}ListNode* mergeTwoList(ListNode* L1,ListNode* L2){ListNode* dummy = new ListNode();ListNode* cur = dummy;while(L1&&L2){if(L1->val < L2->val){cur->next = L1;cur = L1;L1 = L1->next;}else{cur->next = L2;cur = L2;L2 = L2->next;}}cur->next = L1 != nullptr ? L1 : L2;ListNode* res = dummy->next;delete dummy;return res;}
};

方法三、使用优先队列

时间复杂度 O(kn×logk)。空间复杂度 O(k) 。

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {struct Node{ListNode* node_ptr;int val;bool operator<(const Node& rhs) const{return val>rhs.val;}};
public:ListNode* mergeKLists(vector<ListNode*>& lists) {priority_queue<Node> Heap;for(auto& node:lists){if(node){Heap.push({node,node->val});}}ListNode* head = nullptr;ListNode* cur = nullptr;while(!Heap.empty()){if(head == nullptr){head = Heap.top().node_ptr;cur = head;}else{cur->next = Heap.top().node_ptr;cur = cur->next;}Heap.pop();if(cur->next){Heap.push({cur->next,cur->next->val});}}return head;}
};