分治应用--万里挑一找假硬币

文章目录

- 1. 问题描述
- 2. 解题思路
- 3. 代码实现

1. 问题描述

n 个硬币中有1枚是假币，真假币唯一的区别是假币重量轻，如何快速找出假币

2. 解题思路

暴力做法，一个一个的称重，O（n）复杂度
分治思路

将硬币等分成两份，若为奇数，多出一枚，放在天平两边
轻的一边包含假币，若相等，则假币是多出的那一枚
对轻的一边继续上述操作，直到找出假币
复杂度O（log n）

3. 代码实现

/*** @description: n 个硬币中有1枚是假币，假币重量轻，如何快速找出假币* @author: michael ming* @date: 2019/7/6 20:37* @modified by: */
#include <iostream>
#include <ctime>
#include <random>
using namespace std;
int findcoin(int *weight, int left, int right, int &weightimes)
{if(left+1 == right)//只有2枚硬币{weightimes++;//称重比较一次if(weight[left] < weight[right])return left;//返回重量小的位置elsereturn right;}int i, mid, weightsumL, weightsumR;weightsumL = weightsumR = 0;mid = left + (right-left)/2;if((right-left+1)%2 == 0)//偶数枚银币{weightimes++;for(i = left; i <= mid; ++i)weightsumL += weight[i];//计算左边重量(计算机没有天平，只能一个个加)for(i = mid+1; i <= right; ++i)weightsumR += weight[i];//右边重量if(weightsumL > weightsumR)//左边重，假币在右边return findcoin(weight,mid+1,right,weightimes);else if(weightsumL < weightsumR)//假币在左边return findcoin(weight,left,mid,weightimes);else//假币不在两边（偶数枚银币）;//什么都不做，不必再找了}else//奇数枚硬币{weightimes++;for(i = left; i <= mid-1; ++i)weightsumL += weight[i];//计算左边重量for(i = mid+1; i <= right; ++i)weightsumR += weight[i];//右边重量if(weightsumL > weightsumR)//左边重，假币在右边return findcoin(weight,mid+1,right,weightimes);else if(weightsumL < weightsumR)//假币在左边return findcoin(weight,left,mid-1,weightimes);else//两边相等（奇数枚硬币），剩余的那个是假币return mid;}
}
int main()
{srand(unsigned(time(0)));int num, i, weightimes = 0;cout << "请输入硬币总个数:";cin >> num;const int coinNum = num;int *weight = new int [coinNum];for(i = 0; i < coinNum; ++i){weight[i] = 10;}i = rand()%num;weight[i] = 9;  //随机生成假币for(i = 0; i < coinNum; ++i)//打印硬币信息{cout << i + 1 << " 硬币重量: " << weight[i] << endl;}cout << "假硬币是第" << findcoin(weight,0,coinNum-1,weightimes)+1 << "个。" << endl;cout << "共称了" << weightimes << "次，找到假币。";delete[]weight;return 0;
}