A place to hold mainly reading notes, and some technical stuff occasionally. 这里主要是一些读书笔记、感悟;还有部分技术相关的内容。
目录[-]
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
// Note that the way of defining a constructor
public class Solution {
Map<Character, Integer> map = new HashMap<Character, Integer>();
public Solution(){
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);
}
public int romanToInt(String s) {
int sum = 0;
int val1, val2;
for (int i = 0; i < s.length(); i++) {
val1 = map.get(s.charAt(i));
if (i < s.length()-1) {
val2 = map.get(s.charAt(i+1));
if (val1 >= val2) {
sum += val1;
} else {
sum += (val2 - val1);
i += 1;
}
} else {
sum += val1;
}
}
return sum;
}
}
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
// The key is to find the rules of formation of a roman number.
public class Solution {
public String intToRoman(int num) {
int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
String[] numerals = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
StringBuilder result = new StringBuilder();
for (int i = 0; i < values.length; i++) {
while (num >= values[i]) {
num -= values[i];
result.append(numerals[i]);
}
}
return result.toString();
}
}
Reference:http://blog.csdn.net/beiyeqingteng/article/details/8547565