• 【Integer To Roman】cpp


    题目:

    Given an integer, convert it to a roman numeral.

    Input is guaranteed to be within the range from 1 to 3999.

    代码:

    class Solution {
    public:
        string intToRoman(int num) {
            if ( num<1 ) return NULL;
            const int size = 7;
            std::string symbol_ori[size] = {"M","D","C","L","X","V","I"};
            int value_ori[size] = {1000,500,100,50,10,5,1};
            // gain extra symbol and value pair
            std::vector<std::string> symbol_extra;
            std::vector<int> value_extra;
            for ( int i = 0; i < size-1; ++i ){
                symbol_extra.push_back(symbol_ori[i]);
                value_extra.push_back(value_ori[i]);
                if ( !(i & 1) ){
                    symbol_extra.push_back(symbol_ori[i+2]+symbol_ori[i]);
                    value_extra.push_back(value_ori[i]-value_ori[i+2]);
                }
                else{
                    symbol_extra.push_back(symbol_ori[i+1]+symbol_ori[i]);
                    value_extra.push_back(value_ori[i]-value_ori[i+1]);
                }
            }
            symbol_extra.push_back(symbol_ori[size-1]);
            value_extra.push_back(value_ori[size-1]);
    
            std::string result;
            for ( size_t i = 0; i < symbol_extra.size(); ++i )
            {
                int k = num / value_extra[i];
                while ( k )
                {
                    result += symbol_extra[i];
                    k--;
                }
                num = num % value_extra[i];
            }
            return result;
        }
    };

    tips:

    根据罗马数字进位的特殊规则,预先补上不能正常进位的symbol和value,这样代码可以非常consice

    ================================================

    第二次过这道题,重写一遍加深印象。

    class Solution {
    public:
        string intToRoman(int num) {
                        if ( num<1 ) return NULL;
                        const int size = 7;
                        string symbol_ori[size] = {"M","D","C","L","X","V","I"};
                        int value_ori[size] = {1000,500,100,50,10,5,1};
                        vector<string> symbol_extra;
                        vector<int> value_extra;
                        for ( int i=0; i<size-1; ++i ){
                            symbol_extra.push_back(symbol_ori[i]);
                            value_extra.push_back(value_ori[i]);
                            if ( !(1 & i) )
                            {
                                symbol_extra.push_back(symbol_ori[i+2]+symbol_ori[i]);
                                value_extra.push_back(value_ori[i]-value_ori[i+2]);
                            }
                            else
                            {
                                symbol_extra.push_back(symbol_ori[i+1]+symbol_ori[i]);
                                value_extra.push_back(value_ori[i]-value_ori[i+1]);
                            }
                        }
                        symbol_extra.push_back(symbol_ori[size-1]);
                        value_extra.push_back(value_ori[size-1]);
                        string ret = "";
                        for ( int i=0; i<value_extra.size(); ++i )
                        {
                                    int count = num / value_extra[i];
                                    while ( count-->0 ) ret = ret + symbol_extra[i];
                                    num = num % value_extra[i];
                        }
                        return ret;
        }
    };
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  • 原文地址:https://www.cnblogs.com/xbf9xbf/p/4494686.html
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