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社区首页 >专栏 >【LeetCode】 12. Integer to Roman【贪心算法】

【LeetCode】 12. Integer to Roman【贪心算法】

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韩旭051
发布2020-06-23 11:09:47
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发布2020-06-23 11:09:47
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文章被收录于专栏:刷题笔记刷题笔记刷题笔记

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 For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven 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. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3 Output: "III" Example 2:

Input: 4 Output: "IV" Example 3:

Input: 9 Output: "IX" Example 4:

Input: 58 Output: "LVIII" Explanation: L = 50, V = 5, III = 3. Example 5:

Input: 1994 Output: "MCMXCIV" Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/integer-to-roman 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

这道题很贪心

class Solution {
public:
    string intToRoman(int num) {
        vector<int> v={1000,900,500,400,100,90,50,40,10,9,5,4,1};
        vector<string> s={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
        string n;
        int i=0;
        while(num>0){
            while(num>=v[i]){
                n+=s[i];
                num-=v[i];
            }
            i++;
        }return n;
    }

};

最快的老哥和这个有什么区别?

把字符串 n+=s[i];换成n.append(s[i]);就快了不少

class Solution 
{
public:
    string intToRoman(int num) 
    {
        vector<int> vecArab = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        vector<string> vecRoma = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};

        std::string strRoman;
        for(int i = 0; i < static_cast<int>(vecArab.size()); ++i)
        {
            while(num >= vecArab[i])
            {
                strRoman.append(vecRoma[i]);
                num -= vecArab[i];
            }
        }
        
        return strRoman;
    }
};
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