标签:style blog color io ar for div sp on
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
搜了一下,才知道这道题可以写这么简洁。
1 class Solution { 2 public: 3 int romanToInt(string s) { 4 if (s.empty()) return 0; 5 int dict[100]; 6 dict[‘I‘] = 1; dict[‘V‘] = 5; dict[‘X‘] = 10; dict[‘L‘] = 50; 7 dict[‘C‘] = 100; dict[‘D‘] = 500; dict[‘M‘] = 1000; 8 9 int ans = dict[s[0]]; 10 for (int i = 1; i < s.length(); ++i) { 11 if (dict[s[i]] <= dict[s[i - 1]]) { 12 ans += dict[s[i]]; 13 } else { 14 ans = ans - 2 * dict[s[i - 1]] + dict[s[i]]; 15 } 16 } 17 return ans; 18 } 19 };
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
1 class Solution { 2 public: 3 string intToRoman(int num) { 4 string str[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}; 5 int value[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}; 6 7 string ret = ""; 8 for (int i = 0; i < 13; ++i) { 9 while (num >= value[i]) { 10 ret += str[i]; 11 num -= value[i]; 12 } 13 } 14 return ret; 15 } 16 };
Leetcode | Roman to Integer & Integer to Roman
标签:style blog color io ar for div sp on
原文地址:http://www.cnblogs.com/linyx/p/3986686.html