zoukankan      html  css  js  c++  java
  • #Leetcode# 13. Roman to Integer

    https://leetcode.com/problems/roman-to-integer/description/

    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.

    代码:

    class Solution {
    public:
        int romanToInt(string s) {
            int val[10] = {1, 5, 10, 50, 100, 500, 1000};
            map<char, int> mp;
            mp['I'] = 1; mp['V'] = 5; mp['X'] = 10; mp['L'] = 50;
            mp['C'] = 100; mp['D'] = 500; mp['M'] = 1000;
            int len = s.length();
            int ans = mp[s[len - 1]];
            int cnt = mp[s[len - 1]];
            for(int i = len - 2; i >= 0; i --) {
                if(mp[s[i]] < cnt)
                    ans = ans - mp[s[i]];
                else ans += mp[s[i]];
                cnt = mp[s[i]];
            }
            return ans;
        }
    };
    

      

  • 相关阅读:
    图片处理连环画特效
    卡片翻页算法
    android 自定义属性
    android 中捕获全局异常
    c++ 学习笔记
    图片怀旧特效处理
    Linux 网络配置
    指针参数传递
    python 读写文件
    PopupWindow 点击外面取消
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9898497.html
Copyright © 2011-2022 走看看