zoukankan      html  css  js  c++  java
  • 13. Roman to Integer (JAVA)

    Roman numerals are represented by seven different symbols: IVXLCD and M.

    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:

    • I can be placed before V (5) and X (10) to make 4 and 9. 
    • X can be placed before L (50) and C (100) to make 40 and 90. 
    • C can be placed before D (500) and M (1000) to make 400 and 900.

    Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

    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.
    class Solution {
        public int romanToInt(String s) {
            String sym = "MDCLXVI";
            int[] val = {1000,500,100,50,10,5,1};
            int num = 0;
            int p; //array pointer
            for(int i = 0; i < s.length(); i++){
                p = sym.indexOf(s.charAt(i));
                if(i+1 < s.length() && sym.indexOf(s.charAt(i+1)) < p){
                    num = num + val[sym.indexOf(s.charAt(i+1))] - val[p];
                    i++;
                }
                else{
                    num += val[p];
                }
            }
            return num;
        }
    }

    解题思路:要处理的一个特殊情况是4,9,40...这类数,所以我们要扫描到当前位之后的那一位,判断是不是这种情况,做相应处理。

  • 相关阅读:
    Gym-102040B Counting Inversion
    hdu 6899 Xor
    CSPS2019游记
    [网络流系列]网络流基础
    [线段树系列]几道不错的线段树题目题解
    浅谈矩阵[简洁易懂]——上篇
    DP动态规划学习笔记——高级篇上
    DP动态规划学习笔记
    [点分治系列] 静态点分
    [数论系列] 素数篇
  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/10769172.html
Copyright © 2011-2022 走看看