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:
    继承和多态
    访问限制
    返回函数
    类和实例
    requests
    函数的参数
    代码块的快速放置
    19进阶、基于TSP的直流电机控制设计
    18进阶、TLC语言
    17高级、Simulink代码生成技术详解
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9898497.html
Copyright © 2011-2022 走看看