【leetcode】【单链表,vector,queue】【23】Merge k Sorted Lists

#include<iostream>
#include<vector>
#include<queue>
using namespace std;

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
	ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
		ListNode* temp = new ListNode(0);
		ListNode* cur = temp;
		while (l1&&l2){
			if (l1->val < l2->val){
				cur->next = l1;
				l1 = l1->next;
			}
			else{
				cur->next = l2;
				l2 = l2->next;
			}
			cur = cur->next;
		}
		if (l1)
			cur->next = l1;
		else
			cur->next = l2;
		cur = temp->next;
		temp->next = NULL;

		delete temp;
		return cur;
	}

	ListNode* mergeKLists(vector<ListNode*>& lists) {
		int numOfLists = lists.size(); 
		queue<ListNode*> que;
		for (int i = 0; i < numOfLists; ++i){
			if (lists[i])
				que.push(lists[i]);
		}

		if (que.size() == 0)//lists为空或存的全是空链表
			return NULL;

		while (que.size()>1){
			ListNode* l1 = que.front();
			que.pop();
			ListNode* l2 = que.front();
			que.pop();
			que.push(mergeTwoLists(l1, l2));
		}
		return que.front();
	}
	ListNode* createList(ListNode* head){
		int numOfNode;
		int value;
		cout << "please input number of listNode:";
		cin >> numOfNode;
		cin >> value;
		head = new ListNode(value);
		ListNode* cur = head;
		for (int i = 1; i < numOfNode; ++i){
			cin >> value;
			ListNode* temp = new ListNode(value);
			cur->next = temp;
			cur = temp;
		}
		return head;
	}
	void printNode(ListNode* head){
		ListNode* cur = head;
		while (cur){
			cout << cur->val << " ";
			cur = cur->next;
		}
		cout << endl;
	}
};

int main(){
	vector<ListNode*> lists;
	ListNode* head = NULL;
	Solution solution;
	int numOfLists;
	cout << "请输入链表个数:";
	cin >> numOfLists;
	for (int i = 0; i < numOfLists; ++i){
		head = solution.createList(head);
		solution.printNode(head);
		lists.push_back(head);
	}
	
	head = solution.mergeKLists(lists);
	solution.printNode(head);

	system("pause");
	return 0;
}

原文地址:https://www.cnblogs.com/ruan875417/p/4558311.html