LeetCode - Roman to Integer
Problem description
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
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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:
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Input: "III"
Output: 3
Example 2:
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Input: "IV"
Output: 4
Example 3:
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Input: "IX"
Output: 9
Example 4:
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Input: "LVIII"
Output: 58
Explanation: <code>L</code> = 50, <code>V</code> = 5, III = 3.
Example 5:
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Input: "MCMXCIV"
Output: 1994
Explanation: <code>M</code> = 1000, CM = 900, XC = 90 and IV = 4.
Analysis
This is a simple question, here’s the code:
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public class RomantoInteger {
public int romanToInt(String s) {
HashMap<Character, Integer> map = new HashMap<>();
map.put('M', 1000);
map.put('D', 500);
map.put('C', 100);
map.put('L', 50);
map.put('X', 10);
map.put('V', 5);
map.put('I', 1);
int pre = -1;
int res = 0;
for(char c:s.toCharArray()){
int v = map.get(c);
if (v > pre && pre != -1){
res -= 2* pre;
}
res += v;
pre = v;
}
return res;
}
}
An ugly but quick solution from LeetCode is:
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class Solution {
public int romanToInt(String s) {
// Map<String, Integer> romanNumberalsToIntegerMap = new HashMap<>();
// romanNumberalsToIntegerMap.put("I", 1);
// romanNumberalsToIntegerMap.put("IV", 4);
// romanNumberalsToIntegerMap.put("IX", 9);
// romanNumberalsToIntegerMap.put("X", 10);
// romanNumberalsToIntegerMap.put("XL", 40);
// romanNumberalsToIntegerMap.put("XC", 90);
// romanNumberalsToIntegerMap.put("C", 100);
// romanNumberalsToIntegerMap.put("CD", 400);
// romanNumberalsToIntegerMap.put("CM", 900);
// romanNumberalsToIntegerMap.put("M", 1000);
// romanNumberalsToIntegerMap.put("D", 500);
// romanNumberalsToIntegerMap.put("L", 50);
// romanNumberalsToIntegerMap.put("V", 5);
final char[] cArray = s.toCharArray();
int rst = 0;
for(int i = 0; i < cArray.length; i++) {
final char c = cArray[i];
if (c == 'I') {
// 可能需要两个字母代表一个数字
if (i+1 < cArray.length && cArray[i+1] == 'V') {
rst += 4;
i++;
} else if(i+1 < cArray.length && cArray[i+1] == 'X') {
// IX
rst += 9;
i++;
} else {
rst += 1;
}
} else if (c == 'X') {
// 可能需要两个字母代表一个数字
if (i+1 < cArray.length && cArray[i+1] == 'L') {
rst += 40;
i++;
} else if(i+1 < cArray.length && cArray[i+1] == 'C') {
// IX
rst += 90;
i++;
} else {
rst += 10;
}
} else if (c == 'C') {
// 可能需要两个字母代表一个数字
if (i+1 < cArray.length && cArray[i+1] == 'D') {
rst += 400;
i++;
} else if(i+1 < cArray.length && cArray[i+1] == 'M') {
// IX
rst += 900;
i++;
} else {
rst += 100;
}
} else if (c == 'M'){
// 只用当前字母就可以代表数字
rst += 1000;
} else if (c == 'D'){
// 只用当前字母就可以代表数字
rst += 500;
} else if (c == 'L'){
// 只用当前字母就可以代表数字
rst += 50;
} else if (c == 'V'){
// 只用当前字母就可以代表数字
rst += 5;
}
}
return rst;
}
}