1. Leetcode 13. Roman to Integer

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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

Difficulty:Easy

Category:Array


2. Analyze

我们需要用到map数据结构,来将罗马数字的字母转化为对应的整数值,因为输入的一定是罗马数字,那么我们只要考虑两种情况即可: 第一,如果当前数字是最后一个数字,或者之后的数字比它小的话,则加上当前数字 第二,其他情况则减去这个数字

3. Solution

class Solution {
 public:
  int romanToInt(string s) {
    int res = 0;
    unordered_map<char, int> value{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
    for (int i = 0; i < s.size(); ++i) {
      int val = value[s[i]];
      if (i == s.size() - 1 || value[s[i + 1]] <= value[s[i]])
        res += val;
      else
        res -= val;
    }
    return res;
  }
};

我们也可以每次跟前面的数字比较,如果小于等于前面的数字,我们先加上当前的数字,如果大于的前面的数字,我们加上当前的数字减去二倍前面的数字,这样可以把在上一个循环多加数减掉,参见代码如下:

class Solution {
 public:
  int romanToInt(string s) {
    int res = 0;
    unordered_map<char, int> m{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
    for (int i = 0; i < s.size(); ++i) {
      if (i == 0 || m[s[i]] <= m[s[i - 1]])
        res += m[s[i]];
      else
        res += m[s[i]] - 2 * m[s[i - 1]];
    }
    return res;
  }
};
By guozetang            Updated: 2020-09-19 13:02:30

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