• 面试经典150题——Day17


    一、题目

    13. Roman to Integer

    Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

    Symbol Value
    I 1
    V 5
    X 10
    L 50
    C 100
    D 500
    M 1000
    For example, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 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.

    Example 1:

    Input: s = “III”
    Output: 3
    Explanation: III = 3.
    Example 2:

    Input: s = “LVIII”
    Output: 58
    Explanation: L = 50, V= 5, III = 3.
    Example 3:

    Input: s = “MCMXCIV”
    Output: 1994
    Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

    Constraints:

    1 <= s.length <= 15
    s contains only the characters (‘I’, ‘V’, ‘X’, ‘L’, ‘C’, ‘D’, ‘M’).
    It is guaranteed that s is a valid roman numeral in the range [1, 3999].

    题目来源: leetcode

    二、题解

    如果小值在大值前面,就减去该值,如果大值在小值前面,直接加上该值

    class Solution {
    public:
        int romanToInt(string s) {
            int n = s.length();
            unordered_map<char,int> map;
            map['I'] = 1;
            map['V'] = 5;
            map['X'] = 10;
            map['L'] = 50;
            map['C'] = 100;
            map['D'] = 500;
            map['M'] = 1000;
            int res = 0;
            for(int i = 0;i < n;i++){
                if(i < n - 1 && map[s[i]] < map[s[i+1]]) res -= map[s[i]];
                else res += map[s[i]];
            }
            return res;
        }
    };
    
    • 1
    • 2
    • 3
    • 4
    • 5
    • 6
    • 7
    • 8
    • 9
    • 10
    • 11
    • 12
    • 13
    • 14
    • 15
    • 16
    • 17
    • 18
    • 19
    • 20
  • 相关阅读:
    微服务架构的可观察性设计模式
    浅谈数据中心机房UPS蓄电池在线监测系统研究
    milvus数据库搜索
    java面向对象的内存分析
    buffers与cached的异同
    【深度学习】GPT-3
    【Log日志】springboot项目中集成Log日志详解
    有趣的按钮分享
    gulp打包vue3+jsx+less插件
    【剑指offer|图解|链表】链表的中间结点 + 链表中倒数第k个结点
  • 原文地址:https://blog.csdn.net/weixin_46841376/article/details/133855570