[LeetCode] 560. Subarray Sum Equals K

题：https://leetcode.com/submissions/detail/174200311/

题目

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, 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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

思路

题目大意

将 罗马数字 转化 十进制数。
1.罗马数字有7种符号，每种代表不同的数字，通过加和得出最终的 nums。
2.两个 罗马数字若 前者比后者小，那么 表示的数 是 后数 减 前数。

解题思路

1.逆向计算

若从 s的逆向看，若prevalue<=value，res += value；否则 res -= value.

2.正向计算

对于 s[i] ，只有 s[i] >=s[i+1]，res += s[i] ；否则 res -= s[i]。
最后 res + s[-1]

code

1.逆向计算

class Solution:
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """

        numsdict = {

            'I': 1,
            'V': 5,
            'X': 10,
            'L': 50,
            'C': 100,
            'D': 500,
            'M': 1000,
        }
        res = 0
        prevalue = 0

        for i in range(len(s)-1,-1,-1):
            value = numsdict[s[i]]
            if value >= prevalue:
                res +=  value
            else:
                res -= value
            prevalue = value

        return res

2.正向计算

class Solution:
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """

        numsdict = {

            'I': 1,
            'V': 5,
            'X': 10,
            'L': 50,
            'C': 100,
            'D': 500,
            'M': 1000,
        }
        res = 0

        for i in range(0,len(s)-1):

            if numsdict[s[i]]<numsdict[s[i+1]]:
                res -= numsdict[s[i]]
            else:
                res += numsdict[s[i]]
        return res + numsdict[s[len(s)-1]]