求最大公约数

1.暴力枚举，试图找一个合适的整数i，看看这个整数能否被两个整数numberA和numberB同时整除。i从2开始累加，一直加到numberA和numberB中较小参数的一半。循环结束后，上一次找到的能够被两整数的最大i值，就是两数的最大公约数。

package 最大公约数;

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int NumberA = sc.nextInt();
		int NumberB = sc.nextInt();
		System.out.println(gcd(NumberA,NumberB));
	}
	//暴力枚举，试图找一个合适的整数i，看看这个整数能否被两个整数numberA和numberB同时整除。i从2开始累加，
	//一直加到numberA和numberB中较小参数的一半。循环结束后，上一次找到的能够被两整数的最大i值，就是两数的最大公约数。
	private static int gcd(int numberA, int numberB) {
		int smallNumber = numberA<numberB ? numberA:numberB;
		int bigNumber = numberA<=numberB ? numberB:numberA;
		if(bigNumber%smallNumber==0) {
			return smallNumber;
		}
		int greatestCommonDivisor = 1;
		for(int i = 2; i <= smallNumber/2;i++) {
			if(numberA%i==0&&numberB%i==0)
				greatestCommonDivisor = i;
		}
		return greatestCommonDivisor;
				
	}
}

缺点：当输入的两个数分别为100000和100001时，要进行循环10000/2-1 = 4999 次，效率跟不上。

2.辗转相除法

package 最大公约数;

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int NumberA = sc.nextInt();
		int NumberB = sc.nextInt();
		System.out.println(gcd(NumberA,NumberB));
	}

	private static int gcd(int numberA, int numberB) {
		if(numberA%numberB==0)
			return numberB;
		else
			return gcd(numberB,numberA%numberB);
	}
	

3.更相减损术

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int NumberA = sc.nextInt();
		int NumberB = sc.nextInt();
		System.out.println(gcd(NumberA,NumberB));
	}

	private static int gcd(int numberA, int numberB) {
		if(numberA == numberB)
			return numberA;
		if(numberA<numberB) 
			return gcd(numberB-numberA,numberA);
		else
			return gcd(numberA-numberB,numberB);	
	}

	
	
	

1.暴力枚举法：时间复杂度O(min(a,b));

2.辗转相除法：时间复杂度O(log(max(a,b)));

3.更相减损术：时间复杂度O（max（a,b));