Algorithm Analysis and Design: Coin Redemption

Table of contents

Problem Description

Algorithm Design Ideas

 Code

Algorithm running results and calculation time complexity analysis

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Problem Description:

There are several coins of 1 cent, 2 cents, 5 cents, 10 cents, 50 cents, and 100 cents. Now you want to use these coins to pay W yuan. Please design a greedy algorithm to complete it with as few coins as possible . payment behavior. There are several coins of 1 cent, 2 cents, 5 cents, 10 cents, 50 cents, and 100 cents. Now you want to use these coins to pay W yuan. Please design a greedy algorithm to complete it with as few coins as possible . payment behavior.

Algorithm Design Ideas:

Design a for loop to perform operations, and the number of loops is actually linked to the number of sheets. Set the condition, if ww>arr[i] (the largest denomination currently accepted), i-- will not be executed, maybe the denomination can be reduced once. If it cannot be reduced, execute i--, let the next smaller denomination try, and so on. Finally, if ww=0, the result can be output.

 Code:

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			double w = sc.nextDouble();
			int ww = (int)(w*100);//把元单位转换为分
			int arr[] = new int[6];
			for (int i = 0; i < 6; i++) {//输入硬币的几种面额
				int a = sc.nextInt();
				arr[i] = a;
			}
//		Arrays.sort(arr);//若是乱序，则需要排序
			int num = 0;//存储使用张数
			for (int i = 5; i >= 0;) {//因为排序默认从小到大，所以这里要从后往前
				if (ww >= arr[i]) {
					ww = ww - arr[i];
//				i++;
					num++;//每使用一张，加1
				} else {
					i--;//把i--安排在这里，上面只要最大的面额还能用，就不减；不能用了，就执行i--
				}
				if (ww == 0) {
					System.out.println(num);//输出张数
					break;
				}
			}
		}
	}
}

Algorithm running results and calculation time complexity analysis:

If sorting is required, the time complexity is T(n)=O(nlogn)

If no sorting is required, the time complexity is T(n)=O(n)