LeetCode#12 Integer to Roman && #13* Roman to Integer(罗马数字转换)
2017-07-17 10:34
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12题:
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
13题:
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
这两道题就是说将int类型的整数转成罗马数字或者是将罗马数字转成int类型的整数。
【罗马数字】
1~9: {“I”, “II”, “III”, “IV”, “V”, “VI”, “VII”, “VIII”, “IX”};
10~90: {“X”, “XX”, “XXX”, “XL”, “L”, “LX”, “LXX”, “LXXX”, “XC”};
100~900: {“C”, “CC”, “CCC”, “CD”, “D”, “DC”, “DCC”, “DCCC”, “CM”};
1000~3000: {“M”, “MM”, “MMM”}.
【整数转罗马数字】
就是一个简单的模拟,技巧:高位罗马数字串加在已有的数字串的前面,可以使用string的操作(即 ans = roman[][]+ans).
【整数转罗马数字】
这个题要是不观察规律的话,模拟起来就很麻烦!!
观察到罗马数字和Integer的转换的小规律是:
IV = 5 - 1 = (-1) + 5 = 4
VI = 5 + 1 = 5 + 1 = 6
I在V前面,由于I比V小,所以I被解释为负数。
V在I后面,由于V比I大,所以V给解释为整数。
继续看几个例子。
VII = 5 + 1 + 1 = 7
IX = (-1) + 10 = 9
所以,可以一次把输入字符串扫描一遍,从第一个字符开始,到最后一个字符为止,一次比较当前字符a和当前字符的下一个字符b。如果a< b ,解释为 b - a,否则如果a >= b, 解释为a + b。那实际上到这里就很好解决了。
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
13题:
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
这两道题就是说将int类型的整数转成罗马数字或者是将罗马数字转成int类型的整数。
【罗马数字】
1~9: {“I”, “II”, “III”, “IV”, “V”, “VI”, “VII”, “VIII”, “IX”};
10~90: {“X”, “XX”, “XXX”, “XL”, “L”, “LX”, “LXX”, “LXXX”, “XC”};
100~900: {“C”, “CC”, “CCC”, “CD”, “D”, “DC”, “DCC”, “DCCC”, “CM”};
1000~3000: {“M”, “MM”, “MMM”}.
【整数转罗马数字】
就是一个简单的模拟,技巧:高位罗马数字串加在已有的数字串的前面,可以使用string的操作(即 ans = roman[][]+ans).
class Solution { public: string intToRoman(int num) { string roman[][10]= { {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"}, {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"}, {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"}, {"", "M", "MM", "MMM"} }; string ans; int index=0; while(num>0) { int a=num%10; ans=roman[index][a]+ans; index++; num=num/10; } return ans; } };
【整数转罗马数字】
这个题要是不观察规律的话,模拟起来就很麻烦!!
观察到罗马数字和Integer的转换的小规律是:
IV = 5 - 1 = (-1) + 5 = 4
VI = 5 + 1 = 5 + 1 = 6
I在V前面,由于I比V小,所以I被解释为负数。
V在I后面,由于V比I大,所以V给解释为整数。
继续看几个例子。
VII = 5 + 1 + 1 = 7
IX = (-1) + 10 = 9
所以,可以一次把输入字符串扫描一遍,从第一个字符开始,到最后一个字符为止,一次比较当前字符a和当前字符的下一个字符b。如果a< b ,解释为 b - a,否则如果a >= b, 解释为a + b。那实际上到这里就很好解决了。
class Solution { public: int romanToInt(string s) { int i; int ans=GetNum(s[0]); for(i=1;i<s.size();i++) { if(GetNum(s[i])>GetNum(s[i-1])) ans+=GetNum(s[i])-2*GetNum(s[i-1]); else ans+=GetNum(s[i]); } return ans; } int GetNum(char ch) { switch(ch) { case 'I':return 1; case 'V':return 5; case 'X':return 10; case 'L':return 50; case 'C':return 100; case 'D':return 500; case 'M':return 1000; } } };
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