在Java中总结两个巨大的长

Jim*_*hi2 1 java integer integer-overflow biginteger long-integer

我写了这个方法,在arrayList中搜索是否有两个数字,它们总和等于变量elem.问题是变量的总和超过了long类型的维数.我怎么写呢?

public static boolean searchSum(ArrayList<Long> array, long elem) {
   int left = 0, right = array.size()-1;
   while (left<right) {
     long n1=Long.valueOf(array.get(left));
     long n2=Long.valueOf(array.get(right));
     if ((n1+n2)==elem) return true;
     else if ((n1+n2)<elem) left++;
     else right--;
   }
   return false;
 }
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孙兴斌*_*孙兴斌 6

你可以用java.math.BigInteger.它可以存储不可变的任意精度整数.

例如:

public BigInteger sum(Long number1, Long number2) {
    BigInteger bigNumber1 = BigInteger.valueOf(number1);
    BigInteger bigNumber2 = BigInteger.valueOf(number2);
    BigInteger result = bigNumber1.add(bigNumber2);
    return result;
}
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在这种情况下,您可以像这样重写此方法:

public static boolean searchSum(ArrayList<Long> array, long elem) {

    BigInteger bigElem = BigInteger.valueOf(elem);        
    int left = 0, right = array.size() - 1;

    while (left < right) {
        BigInteger n1 = BigInteger.valueOf(array.get(left));
        BigInteger n2 = BigInteger.valueOf(array.get(right));
        BigInteger sum = n1.add(n2);

        if (sum.equals(bigElem)) {
            return true;
        } else if (sum.compareTo(bigElem) < 0) {
            left++;
        } else {
            right--;
        }
    }
    return false;
}
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