• leetcode 12. Integer to Roman(整数转罗马数字)


    Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

    Symbol       Value
    I             1
    V             5
    X             10
    L             50
    C             100
    D             500
    M             1000
    
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    For example, 2 is written as II in Roman numeral, just two one’s added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

    Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

    I can be placed before V (5) and X (10) to make 4 and 9.
    X can be placed before L (50) and C (100) to make 40 and 90.
    C can be placed before D (500) and M (1000) to make 400 and 900.
    Given an integer, convert it to a roman numeral.

    Example 1:

    Input: num = 3
    Output: “III”
    Explanation: 3 is represented as 3 ones.
    Example 2:

    Input: num = 58
    Output: “LVIII”
    Explanation: L = 50, V = 5, III = 3.

    正好和13题反过来,13题是罗马数字转数字。
    罗马数字就那么13个pattern,把整数转成对应的罗马数字。

    思路:

    凭直觉来,
    比如十进制转二进制会从高位到低位,
    这里的十进制转罗马进制只不过是每一位的除数都不一样。

    细数罗马进制的除数也就13种:1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1
    凭直觉的话会先看有多少个1000,然后得到余数再看有多少个900,以此类推。

    public String intToRoman(int num) {
        int[] val = new int[]{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        String[] sym = new String[]{"M", "CM", "D", "CD", "C", "XC", "L", "XL",
                       "X", "IX", "V", "IV", "I"};
        StringBuilder res = new StringBuilder();
        
        for(int i = 0; i < 13; i++) {
            int cnt = num / val[i];
            if(cnt == 0) continue;
            
            for(int j = 0; j < cnt; j++) res.append(sym[i]);
            num %= val[i];
        }
        return res.toString();
    }
    
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    不过下面这种直接减,不用乘除法的要快2ms

    public String intToRoman(int num) {
        int[] val = new int[]{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        String[] sym = new String[]{"M", "CM", "D", "CD", "C", "XC", "L", "XL",
                       "X", "IX", "V", "IV", "I"};
        StringBuilder res = new StringBuilder();
        
        while(num > 0) {
            for(int i = 0; i < 13; i++) {
                if(num >= val[i]) {
                    num -= val[i];
                    res.append(sym[i]);
                    break;
                }
            }
        }
        return res.toString();
    }
    
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  • 原文地址:https://blog.csdn.net/level_code/article/details/127421347