下面是四种用java语言编程实现的求最大公约数的方法：

package gcd;import java.util.ArrayList;
import java.util.List;public class gcd {public static void main(String[] args) {long startTime;long endTime;long durationTime;int[] testArray1 = new int[]{784, 988, 460, 732, 548, 998, 672, 1024, 888, 512};int[] testArray2 = new int[]{1024, 82, 92, 128, 58, 2014, 512, 88, 582, 788};for (int i = 0; i < 10; i++) {startTime = System.nanoTime();System.out.println("欧几里得方法：" + Euclid(testArray1[i],testArray2[i]));endTime = System.nanoTime();durationTime = endTime - startTime;System.out.println("欧几里得算法耗时：" + durationTime + "\n");startTime = System.nanoTime();System.out.println("连续整数检測法：" + consecutiveIntegersTest(testArray1[i], testArray2[i]));endTime = System.nanoTime();durationTime = endTime - startTime;System.out.println("连续整数检測算法耗时：" + durationTime + "\n");startTime = System.nanoTime();System.out.println("辗转相减法：" + consecutiveSubstract(testArray1[i], testArray2[i]));endTime = System.nanoTime();durationTime = endTime - startTime;System.out.println("辗转相减算法耗时：" + durationTime + "\n");startTime = System.nanoTime();System.out.println("分解质因数法：" + primeFactors(testArray1[i], testArray2[i]));endTime = System.nanoTime();durationTime = endTime - startTime;System.out.println("分解质因数算法耗时：" + durationTime);}}/*** 欧几里得算法求最大公约数* @param no1* @param no2* @return*/public static int Euclid(int no1, int no2) {int remainder;remainder = no1%no2;while(remainder != 0) {no1 = no2;no2 = remainder;remainder = no1%no2;}return no2;}/*** 连续整数检測法* @param m* @param n* @return*/public static int consecutiveIntegersTest(int m, int n) {int t;if (m > n) t = n;elset = m;while(true) {if (m%t == 0 && n%t == 0)break;elset = t - 1;}return t;}/*** 辗转相减法* @param num1* @param num2* @return*/public static int consecutiveSubstract(int num1, int num2) {while(true) {if (num1 > num2)num1 -= num2;else if (num1 < num2)num2 -= num1;elsereturn num1;}}/*** 分解质因数法* @param primeNum1* @param primeNum2* @return*/public static int primeFactors(int primeNum1, int primeNum2) {int prime_gcd = 1;int compareListSize;int temp1, temp2;int pn1 = primeNum1, pn2 = primeNum2;List<Integer> num1List = new ArrayList<Integer>();List<Integer> num2List = new ArrayList<Integer>();List<Integer> sameNumList = new ArrayList<Integer>();//求出质因数for (int i = 2; i < pn1/2;) {       //注意此处用的是pn1，而不是primeNum1，primeNum1的值在以下的运行过程会不断减小if (primeNum1%i == 0) {       //求余数，假设能被整除，返回truetemp1 = primeNum1 / i;        //求商primeNum1 = temp1;     //将商赋值给primeNum1。又一次推断余数是否为0num1List.add(i);        //将质因数放入num1List} else if (primeNum1%i != 0) {i = i + 1;     //假设余数不等于0。除数i加1，继续求余数}}for (int i = 2; i < pn2/2;) {if (primeNum2%i == 0) {temp2 = primeNum2 / i;primeNum2 = temp2;num2List.add(i);} else if (primeNum2%i != 0) {i = i + 1;}}int num1ListSize = num1List.size();int num2ListSize = num2List.size();if (num1ListSize < num2ListSize) {for (int i = 0; i < num1List.size();) {if (num2List.contains(num1List.get(i))) {prime_gcd *= num1List.get(i);num2List.remove(num2List.indexOf(num1List.get(i)));num1List.remove(i);if (num1List.size() == 0 || num2List.size() == 0)break;} else {i = i + 1;}}} else {for (int i = 0; i < num2List.size(); ) {if (num1List.contains(num2List.get(i))) {prime_gcd *= num2List.get(i);num1List.remove(num1List.indexOf(num2List.get(i)));num2List.remove(i);if (num1List.size() == 0 || num2List.size() == 0)break;} else {i = i + 1;}}}return prime_gcd;}
}