算法每日一题——08.11

题目链接

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Muban {
	public static Map<Integer, Integer> planksMap = new HashMap<>();
	public static int num_4;
	public static int num_2;
	
	public static void main(String[] args)
	{
		int n;
		Scanner inScanner = new Scanner(System.in);
		n = inScanner.nextInt();
		int input;
		num_2 = 0;
		num_4 = 0;
		for (int i = 0; i < n; i++)
		{
			input = inScanner.nextInt();
			extracted(input);
		}
		int operation_number = inScanner.nextInt();
		for (int i = 0; i < operation_number; i++)
		{
			String judge_opsitive = inScanner.next();
			input = inScanner.nextInt();
			if (judge_opsitive.equals("+"))
			{
				extracted(input);
			}
			else 
			{
				int pre_length = planksMap.get(input);
				planksMap.put(input, pre_length - 1);
				if (pre_length % 4 == 0)
				{
					num_2++;
					num_4--;
				}
				else if (pre_length % 4 == 2)
					num_2--;
			}
			if (num_4 >= 2)
				System.out.println("Yes");
			else if (num_4 == 1 && num_2 >= 2)
				System.out.println("Yes");
			else {
				System.out.println("No");
			}
		}
		inScanner.close();
	}

	public static void extracted(int input) {
		if (planksMap.containsKey(input))
		{
			int pre_length = planksMap.get(input);
			planksMap.put(input, pre_length + 1);
			if (pre_length % 4 == 1)
			{
				num_2++;
			}
			else if (pre_length % 4 == 3)
			{
				num_2--;
				num_4++;
			}
		}
		else {
			planksMap.put(input, 1);
		}
	}
}

本来想暴力求解，但是。。。思路是用一个集合存储每个长度出现的次数。再用两个变量存储有几对可以组成正方形的木棍，有几对可以组成长方形的。这里简单的推导一下即可。

题目链接
想了半天没想出来，总是有些出错。
参考博客