实现sum函数

[wwf06802136]

有两个数组，int a[] ={1,2,3}, b[] = {4,5} 得出的结果是123+45 = 168；求实现两个数组相加的
sum（int value1,int value2）方法？

[gesanri]

public class TestSum {
public static void main(String args[]){
int a[] ={1,2,3};
int b[] = {4,5};
System.out.println("The sum is: " + sum(a,b));
}

public static int sum(int a[],int b[]){
String stra = "";
String strb = "";
for(int i = 0; i < a.length; i++)
stra += String.valueOf(a[i]);
for(int i = 0; i < b.length; i++)
strb += String.valueOf(b[i]);
return Integer.parseInt(stra) + Integer.parseInt(strb);
}
}

[gesanri]

public class TestSum {
	public static void main(String args[]){
		int a[] ={1,2,3};
		int b[] = {4,5};
		System.out.println("The sum is: " + sum(a,b));
	}
	
	public static int sum(int a[],int b[]){
		String stra = "";
		String strb = "";
		for(int i = 0; i < a.length; i++)
			stra += String.valueOf(a[i]);
		for(int i = 0; i < b.length; i++)
			strb += String.valueOf(b[i]);
		return Integer.parseInt(stra) + Integer.parseInt(strb);
	}
}

[bigbro001]

另外一种常规思维的方法。。。

public class SumTest
{
  public static void main(String[] args)
  {
    int[] a = {1, 2, 3};
    int[] b = {4, 5};
    System.out.println(sum(a, b));
  }
  
  public static long sum(int[] a, int[] b)
  {
    long A = 0;
    long B = 0;
    for (int i = a.length - 1, j = 0; i >= 0; i--, j++)  
      A += a[i] * Math.pow(10, j);
    for (int i = b.length - 1, j = 0; i >= 0; i--, j++)
      B += b[i] * Math.pow(10, j);
    return (A + B);
  }
}

[zhanghai921]

public class SumTest {
	public static void main(String[] args){
		int[] a ={1,2,3},b={4,5};
		sum(a,b);
	}
	
	public static int sum(int[] a,int[] b){
		StringBuffer sa = new StringBuffer();
		StringBuffer sb = new StringBuffer();
		for(int i=0;i<a.length;i++){
			sa.append(a[i]);
		}
		for(int i =0;i<b.length;i++){
			sb.append(b[i]);
		}
		return Integer.parseInt(sa.toString())+Integer.parseInt(sb.toString());
		
	}
}

[火龙果被占用了]

public class Test1 {

public static void main(String[] args) {
int[] a = { 1, 2, 3 }, b = { 4, 5 };
int k = sum(a, b);
System.out.println(k);
}

public static int sum(int[] num1, int[] num2) {
int j = num1.length - 1;
int k = num2.length - 1;
int up = 0;
int sum = 0;
int mul = 1;
while (j > -1 || k > -1) {
int n = (j < 0 ? 0 : num1[j--]) + (k < 0 ? 0 : num2[k--]) + up;
sum += n % 10 * mul;
up = n / 10;
mul *= 10;
}
return sum;
}
}

[sforiz]

up 1

[zxlion]

up 2

[Leopard_911]

public static int SUM(int a[],int b[]){
int size = (a.length > b.length ? a.length : b.length) ;
int sum = 0 ;
int sub = 0 ; //进位标志
int temp ;
for(int i = size - 1 , k = 0 ; i >= 0 ; i-- , k++){
if( k >= a.length){
if(sub == 1){
sum += (int)(b[i] + 1) * Math.pow(10, size - i - 1) ;
sub = 0 ;
}
else{
sum += (int)(b[i] * Math.pow(10, size - i - 1)) ;
}
}
else if(k >= b.length){
if(sub == 1){
sum += (int) ((a[i] + 1) * Math.pow(10, size - i - 1)) ;
sub = 0 ;
}
else{
sum += (int)(a[i] * Math.pow(10, size - i - 1)) ;
}
}
else{
if(sub == 1){
temp = a[a.length - k - 1] + b[b.length - k - 1] + 1;
sub = 0 ;
}
else{
temp = a[a.length - k - 1] + b[b.length - k - 1] ;
}
if(temp >= 10){
sub = 1 ;
}
sum += (temp % 10) * Math.pow(10, size - i - 1);
}
}
return sum;
}

[followith]

public class SumArray {
public int sum(int[] value1,int[] value2){
return getValue(value1) + getValue(value2);
}

public int getValue(int[] value){
int number = 0;
if(null != value && value.length > 0){
StringBuffer numberBuffer = new StringBuffer();
for(int i = 0; i< value.length;i++){
numberBuffer.append(value[i]);
}
number = Integer.parseInt(numberBuffer.toString());
}
return number;
}
public static void main(String args[]){
int[] value1 = {9,2,4};
int[] value2 = {5,2,0};
SumArray sumArray = new SumArray();
System.out.println(sumArray.sum(value1,value2));
}
}

[wwf06802136]

如果访问时数据量比较大，是否应该转为long类型，或者哪个算法会比较好点~