leetcode 279. Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

// ConsoleApplication1.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include<vector>
#include<iostream>
using namespace std;

class Solution {
private:
	int minnum;
	vector<int>minpath;
public:
	void do_once(int squarenum,int& remain,vector<int>&path)
	{
		if(path.size()>=minnum-1)
			return;
		for(int i=squarenum;i>=1;i--)
		{
			vector<int>pathcopy=path;
			pathcopy.push_back(i);
			int rem=remain-i*i;
			if(rem==0)
			{
				if(pathcopy.size()<minnum)
				{
					minnum=pathcopy.size();
					minpath=pathcopy;
				}
			}
			else if(rem>0)
			{
				if(pathcopy.size()+1<minnum)
				{
					int startnum=sqrt(remain-i*i);
					if(startnum>i)
						startnum=i;
					do_once(startnum,rem,pathcopy);

				}
			}
		}
	}
	int numSquares(int n) {
		int outsquare=sqrt(n);
		minnum=100000;
		while(outsquare!=0)
		{
			if(n/(outsquare*outsquare)>=minnum)
				return minnum;
			int remain=n;
			vector<int>path;
			do_once(outsquare,remain,path);
			outsquare--;
		}
	}
};

int _tmain(int argc, _TCHAR* argv[])
{
	Solution sl;
	cout<<sl.numSquares(7168);

	system("pause");
	return 0;
}

タイムアウト
少し修正する

class Solution {
private:
	int minnum;
	vector<int>minpath;
public:
	void do_once(int squarenum,int& remain,vector<int>&path)
	{
		if(path.size()>=minnum-1)
			return;
		for(int i=squarenum;i>=1;i--)
		{
			vector<int>pathcopy=path;
			pathcopy.push_back(i);
			int rem=remain-i*i;
			if(rem==0)
			{
				if(pathcopy.size()<minnum)
				{
					minnum=pathcopy.size();
					minpath=pathcopy;
				}
			}
			else if(rem>0)
			{
				if(pathcopy.size()+1<minnum)
				{
					int startnum=sqrt(remain-i*i);
					if(startnum>i)
						startnum=i;
					do_once(startnum,rem,pathcopy);
				}
			}
		}
	}
	int numSquares(int n) {
		int outsquare=sqrt(n);
		minnum=100000;
		while(outsquare!=0)
		{
			if(n/(outsquare*outsquare)>=minnum)
				return minnum;
			int insquare=sqrt(n-outsquare*outsquare);
			if(insquare>outsquare)
				insquare=outsquare;
			int remain=n-outsquare*outsquare;
			if(remain==0)
				return 1;
			vector<int>path;
			path.push_back(outsquare);
			do_once(insquare,remain,path);
			outsquare--;
		}
		return minnum;
	}
};

accepted