Roman to Integer

罗马数字转整数的算法解法

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博客围绕罗马数字转整数问题展开，介绍了罗马数字的表示规则，包括一般从左到右从大到小排列，特殊情况左边小于右边时用右边减左边。还给出两种解法，个人解法通过对象存储映射关系循环判断，社区解法在循环时处理逆序对方式不同。

一问题描述

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:

1. I can be placed before V (5) and X (10) to make 4 and 9.
1. X can be placed before L (50) and C (100) to make 40 and 90.
1. C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

翻译：

罗马数字转换成整数。一般情况下，罗马数字从左到右是从大到小。特殊情况下，会有左边的小于右边的，此时用右边一个减去左边一个。这种情况有：

1. I可以在V和X的左边组合成4和9；
1. X可以在L和C的左边组合成40和90；
1. C可以在D和M的左边组合成400和900；

二解法

1. 第一解法（个人）

思路：

用一个对象存储罗马数字和整数的映射关系。然后循环判断，如果左边小于右边，就用右边减去左边然后索引自增2。否则直接相加。

代码:

var romanToInt = function(s) {
    
    let roman = {
        'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000
    }
    let num = 0;
    
    for (let i = 0; i < s.length; i++) {
        if(s[i+1])
            if (roman[s[i]] >= roman[s[i+1]]) {
                num += roman[s[i]];
            } else {
                num += (roman[s[i+1]] - roman[s[i]]);
                i++; 
            }
        else
            num += roman[s[i++]];
    }
    return num;
    
};

结果：

3999 / 3999 test cases passed.
Status: Accepted
Runtime: 140 ms
Memory Usage: 39.9 MB

2. 第二解法（社区）

思路：

与第一解法一样，但在循环时，不将一个逆序对看作一个整体，而是左边小于右边直接将左边的数字和结果整数相减，否则相加，这样省去了判断索引是否是最后一个。

代码：

let map = {
    'M' : 1000, 'D' : 500, 'C' : 100, 'L' : 50,
    'X' : 10, 'V' : 5, 'I' : 1,
};
var romanToInt = function(s) {
  let result = 0

  for (i=0;i<s.length-1;i++){
    if(map[s[i]] < map[s[i+1]]){
      result -= map[s[i]]
    } else{
      result += map[s[i]]
    }
  }
  result += map[s[s.length-1]]
  
  return result
};

结果：

Runtime: 128 ms, faster than 97.10% of JavaScript online submissions for Roman to Integer.
Memory Usage: 39.6 MB, less than 83.75% of JavaScript online submissions for Roman to Integer

By DoubleJan
2019.7.8