12. Integer to Roman**
https://leetcode.com/problems/integer-to-roman/
题目描述
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:
-
Ican be placed beforeV(5) andX(10) to make 4 and 9. -
Xcan be placed beforeL(50) andC(100) to make 40 and 90. -
Ccan be placed beforeD(500) andM(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: L = 50, V = 5, III = 3.
Example 5:
Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
C++ 实现 1
这道题最直观的思路其实是:
class Solution {
public:
string intToRoman(int num) {
string res;
while (num > 0) {
if (num >= 1000) {
res += "M"
num -= 1000;
} else if (num >= 900) {
res += "CM";
num -= 900;
} else if (num >= 500) {
res += "D";
num -= 500;
}
....
}
}
};
但是写成上面这种形式, 浪费时间和精力. 使用二分法是个不错的选择.
class Solution {
public:
string intToRoman(int num) {
vector<int> choices = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
vector<string> symbols = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"};
string res;
while (num > 0) {
// 找到 choices 中第一个大于 num 的数
auto idx = std::upper_bound(choices.begin(), choices.end(), num);
num -= choices[idx - choices.begin() - 1];
res += symbols[idx - choices.begin() - 1];
}
return res;
}
};