Java整數(shù)相加溢出怎么辦？Java 8 還是厲害！

問題

在之前刷題的時候遇見一個問題，需要解決int相加后怎么判斷是否溢出，如果溢出就返回Integer.MAX_VALUE

解決方案

JDK8已經(jīng)幫我們實(shí)現(xiàn)了Math下，不得不說這個方法是在StackOverflow找到了的，確實(shí)比國內(nèi)一些論壇好多了

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;
}

減法

?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是不一樣的

??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;
}

如何使用？

直接調(diào)用是最方便的，但是為了追求速度，應(yīng)該修改一下，理解判斷思路，因為異常是十分耗時的操作，無腦異常有可能超時。

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