I'm playing around with numbers, and I'm looking to write a small program that will return the largest double-digit of an integer that's passed in for example:

I pass in 2215487 it should return 87

I pass in 98765499 It should return 99

I've tried looking at Math.Max, but I don't believe that's what I'm looking for unless I have overlooked it

CodePudding user response：

You can divide by 10 each time and then compare the remainder of dividing the remaining number by 100. For example:

int v = 98765499;
int m = 0;
v = Math.Abs(v); // negate negative numbers so that we can process these too
while (v >= 10) // if you want to accept single digit initial values for V change this to v > 0
{
    m = Math.Max(m, v % 100);
    v /= 10;
}
Console.WriteLine(m);

We use v >= 10 because we don't want to consider single-digit numbers (e.g. an initial value of v = 5).

v % 100 uses the remainder operator to give us the value left over when we divide by 100.

v /= 10 divides v by 10 and stores the result in v.

I've used Math.Abs( ) to make this work with negative numbers too, though you can take this line out if you only care about positive numbers.

At the end, m will either hold 0 (for single digit numbers or zero) or the highest two-digit value.

Example execution flow:

Highest v % 100 is 99.

Try it online

CodePudding user response：

A very simple solution would be as follows:

    var thenumber = 987654990;
    var s = thenumber.ToString();
    var max = 0;
    for (var i = 0; i < s.Length-1; i  ) {
        int d1 = (int)(s[i] - '0');
        int d2 = (int)(s[i 1] - '0');
        max = Math.Max(max, 10*d1  d2);
    }
    Console.WriteLine(max);

Ie, just iterate over the number and calculate the "pair" at the current position (consisting of the current digit d1 and the next one d2) if the pair is greater than the current max, make it the new max ...