找出一堆硬币中较重的一个（java 算法）

最新推荐文章于 2021-07-23 15:10:45 发布

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最新推荐文章于 2021-07-23 15:10:45 发布

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分类专栏：算法 java 文章标签： java Java JAVA java 二分代码硬币算法

本文链接：https://blog.youkuaiyun.com/sunny_forever/article/details/8571276

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本文介绍了一种在一组硬币中找到唯一不同重量硬币的高效算法。该算法利用平衡秤进行比较，适用于硬币数量为奇数或偶数的情况，并确保称重次数为O(logn)。通过递归划分硬币组并比较重量，快速定位目标。

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1.Description

You have n coins (n may be even or odd) such that n−1 coins are of the same weight and one has different weight. You have a balance scale; you can put any number of coins on each side of the scale at one time, and it will tell you if the two sides weigh the same, or which side is lighter if they do not weigh the same.Outline an algorithm for finding the coin with different weight for each of the following conditions. The number of weighings using your algorithm in each case should be O(logn). You know that one coin weighs more than the others.

2.Algorithm

(1)If the number of coins is even, split them into two parts. If the number is odd, split them into two parts and one coin.

Get the weight of the two parts. The coin which weighs more than other must lie in the part which weighs more.

In the side which weighs more, recursively call step 1. If the weight of two parts are the same, the special coin is the last remaining one.

3.Implementation

public class Coin {
    public static void findCoin(int coins[],int left, int right) {
    	int mid=(int) Math.ceil((left+right)/2); //get the upper limit
    	int LeftSum=0,RightSum=0;
    	if(left==right)
    	{
    		System.out.println("The special coin is coins["+right+"]");
    	}
    	else if(right-left==1)
    	{
    		if(coins[left]>coins[right])
    		{
    			System.out.println("The special coin is coins["+left+"]");
    		}
    		else if(coins[left]>coins[right])
    		{
    			System.out.println("The special coin is coins["+right+"]");
    		}
    	}
    	else 
    	{
    		for(int i=left;i<mid;i++)
    		{
    			LeftSum+=coins[i]; //iteratively get the weight of the left side coins
    			RightSum+=coins[i+mid-left];
    		}
//if the weight of left side is bigger than right side, the special coin should be in the left side
    	if(LeftSum>RightSum)   
    	{
    		right=mid;
    		findCoin(coins,left,right);
    	}
    	else if(LeftSum<RightSum)
    	{
    		left=mid;
    		findCoin(coins,left,right);
    	} 
//if the weight of left side is equal to right side, the special coin should be the last remaining one
    	else if(LeftSum==RightSum)
    	{
    		System.out.println("The special coin is coins["+right+"]");
    	}
    	}
    }
 
    public static void main(String[] args) {
    	int coins1[]={1,1,1,1,1,1,1,1,1,2};
    	int coins2[]={1,1,1,1,2,1,1,1,1};
    	int left1=0,right1=coins1.length-1;
    	findCoin(coins1,left1,right1);
    	int left2=0,right2=coins1.length-1;
    	findCoin(coins2,left2,right2);
    }
}

Screenshot of the result: