我遇到了一个问题:“如何才能将数字反转为整数而不是字符串?”有人能帮我找出答案吗?反转应该反转数字的小数位数,即使用基数10。
发布于 2010-01-11 16:59:58
这应该可以做到:
int n = 12345;
int left = n;
int rev = 0;
while(Convert.ToBoolean(left)) // instead of left>0 , to reverse signed numbers as well
{
int r = left % 10;
rev = rev * 10 + r;
left = left / 10; //left = Math.floor(left / 10);
}
Console.WriteLine(rev);发布于 2010-01-11 16:51:48
像这样的东西?
public int ReverseInt(int num)
{
int result=0;
while (num>0)
{
result = result*10 + num%10;
num /= 10;
}
return result;
}作为一个粗俗的一行(更新:使用本杰明的评论来缩短它):
num.ToString().Reverse().Aggregate(0, (b, x) => 10 * b + x - '0');一个速度更快的1.25行代码:
public static int ReverseOneLiner(int num)
{
for (int result=0;; result = result * 10 + num % 10, num /= 10) if(num==0) return result;
return 42;
}这不是一行,因为我必须包含return 42;。C#编译器不让我编译,因为它认为没有代码路径返回值。
附注:如果你写了这样的代码,而一个同事抓住了它,那么他/她对你所做的一切都是你应得的。请注意!
编辑:我想知道LINQ一行程序的速度有多慢,所以我使用了以下基准代码:
public static void Bench(Func<int,int> myFunc, int repeat)
{
var R = new System.Random();
var sw = System.Diagnostics.Stopwatch.StartNew();
for (int i = 0; i < repeat; i++)
{
var ignore = myFunc(R.Next());
}
sw.Stop();
Console.WriteLine("Operation took {0}ms", sw.ElapsedMilliseconds);
}结果(10^6个随机数在正int32范围内):
While loop version:
Operation took 279ms
Linq aggregate:
Operation took 984ms发布于 2010-01-11 16:51:32
using System;
public class DoWhileDemo {
public static void Main() {
int num;
int nextdigit;
num = 198;
Console.WriteLine("Number: " + num);
Console.Write("Number in reverse order: ");
do {
nextdigit = num % 10;
Console.Write(nextdigit);
num = num / 10;
} while(num > 0);
Console.WriteLine();
}
}https://stackoverflow.com/questions/2040702
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