927. 三等分

927. 三等分

小学生的超时解法：

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <numeric>
using namespace std;

class Solution {
public:
	bool confine = false;
	vector<int> threeEqualParts(vector<int>& A) 
	{
		vector<int> num = A, ans;
		string te;
		for (int i = 0; i < num.size(); i++)
		{
			te += to_string(num[i]);
		}

		//int target = BinaryToDecimalism(num);
		//int begin = num.size() - 1;
		int count_0 = count(num.begin(), num.end(), 0);
		int count_1 = count(num.begin(), num.end(), 1);
		//A中没有0
		if (count_0==0)
		{
			if (num.size()%3==0)
			{
				int te = num.size() / 3;
				return { te - 1,2 * te };
			}
		}
		//A中有一个0
		else if (count_0==1)
		{
			if (num.back() != 0)
			{
				if (count_1 % 3 == 0)
				{
					int n1, n2;
					auto pos = te.find('0');
					if (pos % 3 == 0)
					{
						n1 = count_1 / 3 - 1;
						n2 = 2 * count_1 / 3;
						if (pos== count_1 / 3)
						{
							n2++;
						}
						else if (pos==0)
						{
							n1++;
							n2++;
						}
						return{ n1,n2 };

					}
				}
			}
		}
		//A中没有1
		else if (count_1 == 0)
		{
			return { 0,2 };
		}
		else 
		{

			int len = 0;
			while (len<te.length()-2)
			{
				while (te[te.length()-len-1]=='0')
				{
					len++;
				}
				string pattern = te.substr(te.length() - len - 1);
				if (judge(pattern,te))
				{
					int pos_ans = te.find(pattern);
					int n1 = pos_ans + pattern.length() - 1;
					int n2 = te.find(pattern, pos_ans + pattern.length())+pattern.length() ;
					return { n1,n2 };
				}
				if (confine)
				{
					break;
				}
				len++;
			}
		}
		return { -1,-1 };
	}
	bool judge(string test,string target)
	{
		int cnt = 0;
		auto pos = target.find(test);
		auto temp1 = pos;
		pos = target.find(test, pos + test.length());
		string te = target.substr(temp1 + test.length(), pos - temp1 - test.length());
		if (pos == string::npos)
		{
			confine = 1;
			return false;
		}
		pos = temp1;
		while (pos!=string::npos)
		{
			//if (pos != 0)
			//{
			//	string temp = target.substr(0, pos);
			//	if (temp.find('1') != string::npos)
			//	{

			//		return false;
			//	}
			//}
			auto temp = pos;
			pos = target.find(test, pos + test.length());
			string te = target.substr(temp + test.length(), pos - temp - test.length());
			if (te.find('1') != string::npos)
			{
				return false;
			}
			cnt++;
			if (cnt > 3)
			{
				return false;
			}
		}
		if (cnt<3)
		{
			return false;
		}
		return true;
	}
	int BinaryToDecimalism(vector<int> test)
	{
		int j = 0, ans = 0;
		for (auto i = test.rbegin(); i !=test.rend(); i++)
		{
			ans += pow(2, j)**i;
		}
		return ans;
	}
};



int main()
{
	

	string S = "010110";
	string name = "alex";
	string typed = "aaleex";
	vector<int> num = { 1,0,1,0,1 };
	int N = 85;
	Solution s;
	s.threeEqualParts(num);
	return 0;
}

 

 

 

 

超时解法的思路是从后往前看，组成一个基本的pattern，然后寻找对比，这样效率太低，难以接受；从程序开始看，二进制数只包含0和1，根据题目的要求，其实不难得到这样的一组规律，在数组A中0的个数大于1 的时候，如果存在三等分的情况，那么每一组数组中1的个数是一样的，并且在数组A的尾部存在连续0的情况下，是要统计0的个数（存到num_0中）；二进制中尾部0是不可忽略的（我就犯了这个错误）；如果是想要将每一二进制的值转换成十进制来比较，在数组A的大小上万的情况下，不论那种类型都是不能直接存储的，而string类型不受限制，我们可以直接比较两个string类型的值是否相等；

 

class Solution {
public:
	bool confine = false;
	int num_0, count_1, count_0;
	vector<int> threeEqualParts(vector<int>& A)
	{
		vector<int> num = A, ans;
		string te;
		for (int i = 0; i < num.size(); i++)
		{
			te += to_string(num[i]);
		}

		//int target = BinaryToDecimalism(num);
		//int begin = num.size() - 1;
		count_0 = count(num.begin(), num.end(), 0);
		count_1 = count(num.begin(), num.end(), 1);
		//A中没有0
		if (count_0 == 0)
		{
			if (num.size() % 3 == 0)
			{
				int te = num.size() / 3;
				return { te - 1,2 * te };
			}
		}
		//A中有一个0
		else if (count_0 == 1)
		{
			if (num.back() != 0)
			{
				if (count_1 % 3 == 0)
				{
					int n1, n2;
					auto pos = te.find('0');
					if (pos % 3 == 0)
					{
						n1 = count_1 / 3 - 1;
						n2 = 2 * count_1 / 3;
						if (pos == count_1 / 3)
						{
							n2++;
						}
						else if (pos == 0)
						{
							n1++;
							n2++;
						}
						return{ n1,n2 };

					}
				}
			}
		}
		//A中没有1
		else if (count_1 == 0)
		{
			return { 0,2 };
		}
		else
		{
			int num_0 = 0, num_1 = 0;
			vector<vector<int>>sums(3);
			//计算数组尾部有多少连续的零
			for (auto i = num.rbegin(); i !=num.rend(); i++)
			{
				if (*i == 0)
					num_0++;
				else
					break;
			}
			int j = 0;
			int begin = distance(num.begin(), find(num.begin(), num.end(), 1));
			for (int i = begin; i < num.size(); i++)
			{
				if (num[i]==1)
				{
					num_1++;
				}
				sums[j].push_back(num[i]);
				if (num_1==count_1/3&&j<2)
				{
					num_1 = 0;
					i++;
					for (int k = 0; k < num_0; k++,i++)
					{
						if (num[i] != 0)
						{
							return { -1, -1 };
						}
						sums[j].push_back(num[i]);
					}
					while (num[i]==0)
					{
						i++;
					}
					j++;
					i--;
				}
			}
			vector<string>ans;
			for (int i = 0; i < 3; i++)
			{
				ans.push_back(tranToString(sums[i]));
			}

			if (ans[1] == ans[2]&&ans[2] == ans[0]&&ans[0]==ans[1])
			{
				int n1 = sums[0].size() - 1 + begin;
				int n2 = num.size() - sums[2].size();
                				while (num[n2-1]==0)
				{
					n2--;
				}
                n2 += num_0;
				return { n1,n2 };
			}
		}
		return { -1,-1 };
	}
	long long BinaryToDecimalism(vector<long long> test)
	{
		int j = 0;
		long long ans = 0;
		for (auto i = test.rbegin(); i != test.rend(); i++)
		{
			ans += pow(2, j)**i;
			j++;
		}
		return ans;
	}
	string tranToString(vector<int>test)
	{
		string te;
		for (auto i = test.begin(); i !=test.end(); i++)
		{
			te += to_string(*i);
		}
		return te;
	}

};