LeetCode: Roman to Integer 解题报告

Roman to Integer
Given a roman numeral, convert it to an integer.

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

SOLUTION 1：

思路：

从后往前遍历罗马数字，如果某个数比前一个数小，则把该数在结果中减掉；
反之，则在结果中加上当前这个数；

 package Algorithms.string;

 public class RomanToInt {
     public int romanToInt(String s) {
         if (s == null) {
             return 0;
         }

         int len = s.length();
         int sum = 0;
         int pre = 0;

         for (int i = len - 1; i >= 0; i--) {
             int cur = romanTable(s.charAt(i));

             if (i == len - 1) {
                 // 如果是在尾部，直接加上当前值
                 sum += cur;
             } else {
                 // 判定当前值是不是比前一个值要小，如果小，则需要减去它
                 if (cur < pre) {
                     sum -= cur;
                 } else {
                     sum += cur;
                 }
             }
             pre = cur;
         }

         return sum;
     }

     public int romanTable(char c) {
         int num = 0;
         switch(c) {
             case 'I':
                 num = 1;
                 break;
             case 'V':
                 num = 5;
                 break;
             case 'X':
                 num = 10;
                 break;
             case 'L':
                 num = 50;
                 break;
             case 'C':
                 num = 100;
                 break;
             case 'D':
                 num = 500;
                 break;
             case 'M':
                 num = 1000;
                 break;
             default:
                 num = 0;
                 break;
         }

         return num;
     }
 }

SOLUTION 2:

除了用函数转换，也可以用map来转换。

 public int romanToInt(String s) {
         if (s == null) {
             return 0;
         }

         // bug 1: forget new.
         HashMap<Character, Integer> map = new HashMap<Character, Integer>();
         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);

         int len = s.length();
         int num = 0;
         for (int i = len - 1; i >= 0; i--) {
             int cur = map.get(s.charAt(i));
             if (i < len - 1 && cur < map.get(s.charAt(i + 1))) {
                 num -= cur;
             } else {
                 num += cur;
             }
         }

         return num;
     }

GITHUB 代码：

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/RomanToInt.java