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.
public int romanToInt(String s) {
int[] nums = new int[s.length()];
for (int i = 0; i < nums.length; i++) {
char cur = s.charAt(i); //此方法返回位于字符串的指定索引处的字符。该字符串的索引从零开始。
if (cur == 'M') nums[i] = 1000; //处理字符串加''
else if(cur == 'D') nums[i] = 500;
else if(cur == 'C') nums[i] = 100;
else if(cur == 'L') nums[i] = 50;
else if(cur == 'X') nums[i] = 10;
else if(cur == 'V') nums[i] = 5;
else if(cur == 'I') nums[i] = 1;
}
int sum = 0;
for(int i = 0; i < nums.length - 1; i++) { // 此处只能到i<nums.length-1,因为下面的程序需要比较num[i+1],如果使用nums.length会出现ArrayOutOfBound
if(nums[i + 1] > nums[i]) { // 如果后面的比前面的大，那么就需要减去前面一位的那个值
sum -= nums[i];
} else {
sum += nums[i];
}
}
return sum + nums[nums.length - 1];//因为前面的程序没有计算到最后一位，所以返回的时候要加上
}
}