demo地址

kotlin字符串与数字数字与字符串转换

 fun stringToInt(){
        var a=3
        var b="3"
        a=b.toInt()
        b=a.toString()
    }

kotlin递归之阶乘

     fun factorials(numbers:Int):Int{
        //5的阶乘=5*4*3*2*1
         //4的阶乘=4*3*2*1
         //3的阶乘=3*2*1
         //2的阶乘=2*1
         //1的阶乘=1
         //可以理解为5的阶乘等于5*4的阶乘
         if(numbers==1){
             return 1
         }else{
             return numbers*factorials(numbers-1)
         }
    }

当传入参数5时的结果
2020-08-21 14:16:09.412 7372-7372/com.example.myapplication I/kotlin: result=120
当传入参数100时的结果
2020-08-21 14:22:41.805 7945-7945/com.example.myapplication I/kotlin: result=0

出现这种情况的原因是因为函数的返回值Int取值范围在-2147483648~2147483647，而100的阶乘超过了这个范围，下面换成Long来测试

    fun factorialsLong(numbers:Long):Long{
        //5的阶乘=5*4*3*2*1
        //4的阶乘=4*3*2*1
        //3的阶乘=3*2*1
        //2的阶乘=2*1
        //1的阶乘=1
        //可以理解为5的阶乘等于5*4的阶乘
        if(numbers==1L){
            return 1L
        }else{
            return numbers*factorialsLong((numbers-1))
        }
    }

2020-08-21 14:22:41.805 7945-7945/com.example.myapplication I/kotlin: result=0

此时Long类型也无法装在下100的阶乘，可以用BigInteger来装载

    fun factorialsBigInteger(numbers:BigInteger):BigInteger{
        //5的阶乘=5*4*3*2*1
        //4的阶乘=4*3*2*1
        //3的阶乘=3*2*1
        //2的阶乘=2*1
        //1的阶乘=1
        //可以理解为5的阶乘等于5*4的阶乘
        if(numbers== BigInteger.ONE){
            return BigInteger.ONE
        }else{
            return numbers*factorialsBigInteger((numbers-BigInteger.ONE))
        }
    }

此时得到结果完美perfect

2020-08-21 14:36:56.635 10121-10121/com.example.myapplication I/kotlin: result=93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000