013. 罗马数字转整数 | Leetcode题解

字符          数值
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

输入:"III"
输出: 3

输入:"IV"
输出: 4

输入:"IX"
输出: 9

输入:"LVIII"
输出: 58
解释: L = 50, V= 5, III = 3.

输入:"MCMXCIV"
输出: 1994
解释: M = 1000, CM = 900, XC = 90, IV = 4.

转换表为：
 字符     数值     进位
  I       1       个位值为1
  V       5       个位值为5
  X       10      十位值为1
  L       50      十位值为5
  C       100     百位值为1
  D       500     百位值为5
  M       1000    千位值为1

例如：4:IV    9:IX
    40:XL   90:XC
   400:CD  900:CM
可知：数值大小 == 大的数字位-小的数字位 ---- 《2》

/**
 * @来源： Javascript中文网 - 前端进阶资源教程 https://www.javascriptc.com/
 * @介绍：一个致力于帮助开发者用代码改变世界为使命的平台，每天都可以在这里找到技术世界的头条内容
 * @param {string} s
 * @return {number}
 */
var romanToInt = function(s) {
  let hash={
      I:1,
      V:5,
      X:10,
      L:50,
      C:100,
      D:500,
      M:1000
  }
  let sum=0
  for(let i=0;i<s.length;i++){
    let res
    let cur=hash[s[i]]
    let next=hash[s[i+1]]
    if(next  && cur < next){
      res=next-cur
      i++
    }else{
      res=cur
    }
    sum+=res
  }
  return sum
};

/**
 * @param {number} num
 * @return {string}
 //作者：Alexer-660
//链接：https://leetcode-cn.com/problems/integer-to-roman/solution/12-zheng-shu-zhuan-luo-ma-shu-zi-by-alexer-660/
 */
var romanToInt = function(s) {
    var hashNum = {
        "I":1,
        "V":5,
        "X":10,
        "L":50,
        "C":100,
        "D":500,
        "M":1000
    }
    var result = 0;
    for(let i = 0;i<s.length;i++){
        hashNum[s[i]] < hashNum[s[i+1]] ? result -= hashNum[s[i]] : result += hashNum[s[i]]
    }
    return result;
};

/**
 作者：iamapig120
链接：https://leetcode-cn.com/problems/integer-to-roman/solution/cyu-yan-js-cong-di-wei-kai-shi-an-wei-qu-zhi-cha-b/
 */
 /**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function(s) {
    let keys = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'],
        values = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1],
        res = 0, idx;
    while(s.length) {
        idx = keys.indexOf(s[0] + s[1]);//先判断是不是符合2个字符的单位
        if(idx >= 0){
            s = s.substring(2, s.length);
        }else{
            idx = keys.indexOf(s[0]);
            s = s.substring(1, s.length);
        }
        res += values[idx];//得到对应的值
    }
    return res;
};

//作者：iamapig120
//链接：https://leetcode-cn.com/problems/integer-to-roman/solution/cyu-yan-js-cong-di-wei-kai-shi-an-wei-qu-zhi-cha-b/
class Solution {
public:
    int change(char a) {
        switch(a)
        {
            case 'I':return 1;break;
            case 'V':return 5;break;
            case 'X':return 10;break;
            case 'L':return 50;break;
            case 'C':return 100;break;
            case 'D':return 500;break;
            case 'M':return 1000;break;
        }
        return 0;
    }
    int romanToInt(string s) {
        int SUM=0,sum=0;
        int i;
        for(i=0;i<s.length();i++)
        {
            int num=change(s[i]);
            if(change(s[i+1])>num)num=-num;
            SUM+=num;
        }
        return SUM;
    }
};

// 作者：junex233
// 链接：https://leetcode-cn.com/problems/roman-to-integer/solution/si-lu-qing-xi-shi-xian-jian-dan-de-fang-fa-by-june/
class Solution {
public:
    int romanToInt(string s) {
        int result=0;
        map<char,int> luomab={
            {'I',1},
            {'V',5},
            {'X',10},
            {'L',50},
            {'C',100},
            {'D', 500},
            {'M', 1000}
        };//初始化哈希表
        for(int i=0;i<s.length();i++)
        {
            if(luomab[s[i]] < luomab[s[i+1]])
                result -= luomab[s[i]];
            else
            {
                result += luomab[s[i]];
            }
        }
        return result;
    }
};

# 作者：z1m
# 链接：https://leetcode-cn.com/problems/roman-to-integer/solution/qing-xi-tu-jie-python3-by-ml-zimingmeng/
class Solution:
    def romanToInt(self, s: str) -> int:
        Roman2Int = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
        Int = 0
        n = len(s)

        for index in range(n - 1):
            if Roman2Int[s[index]] < Roman2Int[s[index + 1]]:
                Int -= Roman2Int[s[index]]
            else:
                Int += Roman2Int[s[index]]

        return Int + Roman2Int[s[-1]]

# 作者：zkk-c
# 链接：https://leetcode-cn.com/problems/integer-to-roman/solution/tan-xin-ha-xi-biao-tu-jie-by-ml-zimingmeng/
class Solution:
    def romanToInt(self, s: str) -> int:
#时间复杂度为O(n),空间复杂度为O(1)
        d = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        result = 0
        for i in range(len(s) - 1):
            if d[s[i]] >= d[s[i + 1]]:
                result += d[s[i]]
            else:
                result -= d[s[i]]
        result += d[s[-1]]
        return result