【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;
    }
};

 

posted on 2015-05-11 15:26  承续缘  阅读(174)  评论(0编辑  收藏  举报

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