刷题23. Merge k Sorted Lists

一、题目说明

这个题目是23. Merge k Sorted Lists，归并k个有序列表生成一个列表。难度为Hard，实际上并不难，我一次提交就对了。

二、我的解答

就是k路归并，思路很简单，实现也不难。

#include<iostream>
#include<vector>
using namespace std;
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
	ListNode* mergeKLists(vector<ListNode*>& lists){
		ListNode dummy(-1);
		ListNode *p,*pResult,*cur;
		if(lists.size()<=0) return NULL;
		for(int t=0;t<lists.size();t++){
			p = lists[t];
			pResult = &dummy;
			while(p !=NULL){
				while(pResult->next!=NULL && pResult->next->val < p->val){
					pResult = pResult->next;
			    }
				cur = new ListNode(p->val);
				cur->next = pResult->next;
				pResult->next = cur;
				p = p->next;
			}
		}
		return dummy.next;
	}
};
int main(){
	Solution s;
	ListNode* l1, *l2, *l3, *p;

	// init l1
	l1 = new ListNode(5);
	p = new ListNode(4);
	p->next = l1;
	l1 = p;
	p = new ListNode(1);
	p->next = l1;
	l1 = p;

	// init l2
	l2 = new ListNode(4);
	p = new ListNode(3);
	p->next = l2;
	l2 = p;
	p = new ListNode(1);
	p->next = l2;
	l2 = p;

	// init l1
	l3 = new ListNode(6);
	p = new ListNode(2);
	p->next = l3;
	l3 = p;

	vector<ListNode*> lists;
	lists.push_back(l1);
	lists.push_back(l2);
	lists.push_back(l3);

	ListNode* r = s.mergeKLists(lists);
	while(r != NULL){
		cout<<r->val<<" ";
		r = r->next;
	}
	return 0;
}

不过，性能一般：

Runtime: 172 ms, faster than 21.46% of C++ online submissions for Merge k Sorted Lists.
Memory Usage: 12.8 MB, less than 5.95% of C++ online submissions for Merge k Sorted Lists.

三、优化措施

上面的实现，之所以性能不足，在于一次归并一个队列，用的是插入排序。其实n路归并，可以用优先级队列priority_queue一次实现的。

class Solution {
	struct CompareNode {
	    bool operator()(ListNode* const & p1, ListNode* const & p2) {
	        // return "true" if "p1" is ordered before "p2", for example:
	        return p1->val > p2->val;
           //Why not p1->val <p2->val; ??
	    }
	};
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {

    	ListNode dummy(0);
    	ListNode* tail=&dummy;

    	priority_queue<ListNode*,vector<ListNode*>,CompareNode> queue;

    	for (vector<ListNode *>::iterator it = lists.begin(); it != lists.end(); ++it){
    		if (*it)
    		queue.push(*it);
    	}
    	while (!queue.empty()){
    		tail->next=queue.top();
    		queue.pop();
    		tail=tail->next;

    		if (tail->next){
    			queue.push(tail->next);
    		}
    	}

    	return dummy.next;
    }
};