Roman to Integer (E)
题目
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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(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.
Example 1:
Input: "III"
Output: 3
Example 2:
Input: "IV"
Output: 4
Example 3:
Input: "IX"
Output: 9
Example 4:
Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 5:
Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
题意
将给定的罗马数字转化为整数。
思路
罗马数字的规律是,除了4、9等这类特殊的数字,所有的大数字都在小数字的左边,当小数排在大数左侧时,说明这个小数是需要被减去的。参照此规律,从右往左遍历字符,如果当前数字比右边的数字(若存在)小,则需要在总和中减去当前数,否则加上当前数。
代码实现
Java
class Solution {
public int romanToInt(String s) {
int ans = 0;
Map map = new HashMap<>();
map.put('I', 1);
map.put('V', 5);
map.put('X', 10);
map.put('L', 50);
map.put('C', 100);
map.put('D', 500);
map.put('M', 1000);
for (int i = s.length() - 1; i >= 0; i--) {
char c = s.charAt(i);
if (i < s.length() - 1 && map.get(c) < map.get(s.charAt(i + 1))) {
ans -= map.get(c);
} else {
ans += map.get(c);
}
}
return ans;
}
}
JavaScript
/**
* @param {string} s
* @return {number}
*/
var romanToInt = function (s) {
let sum = 0
let map = new Map([
['M', 1000],
['D', 500],
['C', 100],
['L', 50],
['X', 10],
['V', 5],
['I', 1],
])
let nums = s.split('').map((c) => map.get(c))
for (let i = 0; i < nums.length; i++) {
let sign = i < nums.length - 1 && nums[i] < nums[i + 1] ? -1 : 1
sum += sign * nums[i]
}
return sum
}