Java - How to format a number to a 2 char string?

后端 未结 3 1887
日久生厌
日久生厌 2021-02-18 12:56

How would I format an int between 0 and 99 to a string which is 2 chars. E.g

4  becomes \"04\"
36 becomes \"36\"

Than

3条回答
  •  说谎
    说谎 (楼主)
    2021-02-18 13:18

    Use String.format(). Example:

    class Test{
        public static void main(String[]args){
            System.out.println(String.format("%02d", 42)); //42
            System.out.println(String.format("%02d",  7)); //07
        }
    }
    

    Edit:

    luiscubal's comment got me thinking, so I figured why not benchmark the three alternatives.

    import java.text.*;
    
    interface Fmt{
        public String f(int x);
    }
    
    class Test{
    
        static NumberFormat formatter = new DecimalFormat("00");
    
        static Fmt f1 = new Fmt(){
            public String f(int x){ return ((x<10)?"0":"") + x; }
            public String toString(){return "f1";}
        };
    
        static Fmt f2 = new Fmt(){
            public String f(int x){ return String.format("%02d", x); }
            public String toString(){return "f2";}
        };
    
        static Fmt f3 = new Fmt(){
            public String f(int x){ return formatter.format(x); }
            public String toString(){return "f3";}
        };
    
        public static void main(String[]args){
    
            Fmt[] fmts = new Fmt[]{f1, f2, f3, f3, f2, f1};
    
            for (int x : new int[]{7, 42, 99}){
    
                String s0 = null;
                for (Fmt fmt : fmts)
                    if (s0==null)
                        s0 = fmt.f(x);
                    else
                        if (!fmt.f(x).equals(s0))
                            System.exit(1);
    
                System.out.printf("%02d\n", x);
    
                for (Fmt fmt : fmts){
                    String s = null;
                    int count = 0;
                    System.gc();
                    long t0 = System.nanoTime();
                    for (int i=0; i<100000; i++){
                        count += fmt.f(x).length();
                    }
                    long t1 = System.nanoTime();
    
                    System.out.printf("    %s:%8.2fms, count=%d\n",
                        fmt, (t1-t0)/1000000.0, count);
                }
    
            }
        }
    
    }
    

    The output I got was:

    07
        f1:   11.28ms, count=200000
        f2:  195.97ms, count=200000
        f3:   45.41ms, count=200000
        f3:   39.67ms, count=200000
        f2:  164.46ms, count=200000
        f1:    6.58ms, count=200000
    42
        f1:    5.25ms, count=200000
        f2:  163.87ms, count=200000
        f3:   42.78ms, count=200000
        f3:   42.45ms, count=200000
        f2:  163.87ms, count=200000
        f1:    5.15ms, count=200000
    99
        f1:    5.83ms, count=200000
        f2:  168.59ms, count=200000
        f3:   42.86ms, count=200000
        f3:   42.96ms, count=200000
        f2:  165.48ms, count=200000
        f1:    5.22ms, count=200000
    

    Turns out - if I tested correctly - that SpeedBirdNine had the most efficient solution, albeit badly presented. So I revise my solution to:

    String answer = ((x<10)?"0":"") + x;
    

    In retrospect I think it makes sense, given that my initial solution incurred the cost of parsing the format string, and I'd assume Anthony's formatter also has some manner of data structure behind it being iterated every time.

    And predictably a lot faster by keeping a String[] containing all the 100 strings, but that's probably overkill...

提交回复
热议问题