# Roman Number & Integer

Problem I: Roman to Integer

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

1. 将字符和整数存入map中；

2. 从后往前扫描，如果当前字符代表的值大于后一个字符的值，则用当前的值减去之前累计的值。e.g. "IV" -> 5 - 1 = 4

public class Solution {
public int romanToInt(String s) {
if (s == null || s.length() == 0) {
return 0;
}

HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
hm.put('I', 1);
hm.put('V', 5);
hm.put('X', 10);
hm.put('L', 50);
hm.put('C', 100);
hm.put('D', 500);
hm.put('M', 1000);

int len = s.length();
int sum = hm.get(s.charAt(len - 1));

for (int i = len - 2; i >= 0; i--) {
char cur = s.charAt(i);
char aft = s.charAt(i + 1);
if (hm.get(cur) < hm.get(aft)) {
sum -= hm.get(cur);
} else {
sum += hm.get(cur);
}
}

return sum;
}
}


Problem II: Integer to Roman

Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

1. 存一个数字到字符的映射表，数字从大到小，包含{1000, 900, 500, 400, 100, 90,
50, 40, 10, 9, 5, 4, 1}；

2. 贪心原则，得到对应的字符串。

public class Solution {
public String intToRoman(int num) {
String res = "";
if (num <= 0) {
return res;
}

int[] nums = {1000, 900, 500, 400, 100, 90,
50, 40, 10, 9, 5, 4, 1};
String[] symbols = {"M", "CM", "D", "CD", "C", "XC",
"L", "XL", "X", "IX", "V", "IV", "I"};

int i = 0;

while (num > 0) {
int j = num / nums[i];
num -= j * nums[i];
for ( ; j > 0; j--) {
res += symbols[i];
}
++i;
}

return res;
}
}


posted @ 2015-08-19 19:48  Chapter  阅读(253)  评论(0编辑  收藏  举报