• LeetCode Integer To Roman & Roman To Int


    罗马数字

     有几条须注意掌握;
    (1)基本数字Ⅰ、X 、C 中的任何一个,自身连用构成数目,或者放在大数的右边连用构成数目,都不能超过三个;放在大数的左边只能用一个。
    (2)不能把基本数字 V 、L 、D 中的任何一个作为小数放在大数的左边采用相减的方法构成数目;放在大数的右边采用相加的方式构成数目,只能使用一个。
    (3)V 和 X 左边的小数字只能用Ⅰ。
    (4)L 和 C 左边的小数字只能用×。
    (5)D 和 M 左 边的小数字只能用 C 。

    【对照举例】
    ·个位数举例
    I, 1 II, 2 III, 3 IV, 4 V, 5 VI, 6 VII, 7 VIII,8 IX, 9

    ·十位数举例
    X, 10 XI, 11 XII, 12 XIII, 13 XIV, 14 XV, 15 XVI, 16 XVII, 17 XVIII, 18 XIX, 19 XX, 20 XXI, 21 XXII, 22 XXIX, 29 XXX, 30 XXXIV, 34 XXXV, 35 XXXIX, 39 XL, 40 L, 50 LI, 51 LV, 55 LX, 60 LXV, 65 LXXX, 80 XC, 90 XCIII, 93 XCV, 95 XCVIII, 98 XCIX, 99

    ·百位数举例
    C, 100 CC, 200 CCC, 300 CD, 400 D, 500 DC,600 DCC, 700 DCCC, 800 CM, 900 CMXCIX,999

    ·千位数举例
    M, 1000 MC, 1100 MCD, 1400 MD, 1500 MDC, 1600 MDCLXVI, 1666 MDCCCLXXXVIII, 1888 MDCCCXCIX, 1899 MCM, 1900 MCMLXXVI, 1976 MCMLXXXIV, 1984 MCMXC, 1990 MM, 2000 MMMCMXCIX, 3999

    另外参考 http://blog.csdn.net/menxu_work/article/details/9147209

    方法一:

        public String intToRoman(int num) {
                    StringBuilder stringBuilder=new StringBuilder();
            int i=num;
            if(num>=1000 && num<4000)
            {
                int j=i/1000;
                while(j-->0)
                {
                    stringBuilder.append('M');
                }
                stringBuilder.append(intToRoman(i%1000));
            }
            else if(num>=900 && num<1000)
            {
                stringBuilder.append("CM"+intToRoman(i%900));
            }
            else if(num>=500 && num<900)
            {
                stringBuilder.append("D"+intToRoman(i%500));
            }
            else if(num>=400 && num<500)
            {
                stringBuilder.append("CD"+intToRoman(i%400));
            }
            else if(num>=100 && num<400)
            {
                int j=i/100;
                while(j-->0)
                {
                    stringBuilder.append('C');
                }
                stringBuilder.append(intToRoman(i%100));
            }
            else if(num>=90 && num<100)
            {
                stringBuilder.append("XC"+intToRoman(i%90));
            }
            else if(num>=50 && num<90)
            {
                stringBuilder.append("L"+intToRoman(i%50));
            }
            else if(num>=40 && num<50)
            {
                stringBuilder.append("XL"+intToRoman(i%40));
            }
            else if(num>=10 && num<40)
            {
                int j=i/10;
                while(j-->0)
                {
                    stringBuilder.append('X');
                }
                stringBuilder.append(intToRoman(i%10));
            }
            else if(num>=9 && num<10)
            {
                stringBuilder.append("IX");
            }
            else if(num>=5 && num<9)
            {
                stringBuilder.append("V"+intToRoman(i%5));
            }
            else if(num>=4 && num<5)
            {
                stringBuilder.append("IV");
            }
            else if(num>=1 && num<4)
            {
                while(i-->0)
                {
                    stringBuilder.append('I');
                }
            }
            return stringBuilder.toString();
        }

    方法二:

        public String intToRoman(int num) {
            int [] vals={1000,900,500,400,100,90,50,40,10,9,5,4,1};
            String[] romans={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
            StringBuilder stringBuilder=new StringBuilder();
            for(int i=0;i<vals.length;i++)
            {
                while(num>=vals[i])
                {
                    num-=vals[i];
                    stringBuilder.append(romans[i]);
                }
            }
            return stringBuilder.toString();
        }

    Roman To Int

            int [] vals={1000,900,500,400,100,90,50,40,10,9,5,4,1};
            String[] romans={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
            int result=0,offset=0;
            while(offset<s.length())
            {
                for(int i=0;i<romans.length;i++)
                {
                    if(s.startsWith(romans[i],offset))
                    {
                        result+=vals[i];
                        offset+=romans[i].length();
                        break;
                    }
                }
            }
            return result;
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  • 原文地址:https://www.cnblogs.com/maydow/p/4628291.html
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