StackOverflow热帖：Java整数相加重叠怎么办？Java8一步搞定〜

会呼吸的Coder

发布于 2020-11-03 10:59:02

3110

发布于 2020-11-03 10:59:02

文章被收录于专栏：会呼吸的Coder会呼吸的Coder

＃问题

在之前刷题的时候遇见一个问题，需要解决int相加后如何判断是否重叠，如果重叠就返回Integer.MAX_VALUE

＃解决方案

JDK8已经帮我们实现了Math下，不得不说这个方法是在StackOverflow找到了的，确实比国内一些论坛好多了〜

加法

public static int addExact(int x, int y) {
        int r = x + y;
        // HD 2-12 Overflow iff both arguments have the opposite sign of the result
        if (((x ^ r) & (y ^ r)) < 0) {
            throw new ArithmeticException("integer overflow");
        }
        return r;
}

减法

 public static int subtractExact(int x, int y) {
        int r = x - y;
        // HD 2-12 Overflow iff the arguments have different signs and
        // the sign of the result is different than the sign of x
        if (((x ^ y) & (x ^ r)) < 0) {
            throw new ArithmeticException("integer overflow");
        }
        return r;
}

乘法

public static int multiplyExact(int x, int y) {
        long r = (long)x * (long)y;
        if ((int)r != r) {
            throw new ArithmeticException("integer overflow");
        }
        return (int)r;
}

注意long和int是不一样的

  public static long multiplyExact(long x, long y) {
        long r = x * y;
        long ax = Math.abs(x);

long ay = Math.abs（y）;

        if (((ax | ay) >>> 31 != 0)) {
            // Some bits greater than 2^31 that might cause overflow
            // Check the result using the divide operator
            // and check for the special case of Long.MIN_VALUE * -1
           if (((y != 0) && (r / y != x)) ||
               (x == Long.MIN_VALUE && y == -1)) {
                throw new ArithmeticException("long overflow");
            }
        }
        return r;
}