思想：
首先将字符转换成其对应的ASCII码，然后将ASCII码+1求余65得到余数（65是大写A的ASCII）,最后将余数乘以26的次方（次方数是多少由其所在位置决定）
class Solution: def titleToNumber(self, s): """ :type s: str :rtype: int """ all = 0 for i in range(0,len(s)): all += (ord(s[i])+1)%65*26**(len(s)-1-i) return all下面是大佬的思想：
首先定义一个字典，存放这单个字符对应的数值。然后通过reversed(s)，实现字符串的反转并变成list类型。然后通过letters[c]取出该字符有对应的数值并乘以26的次方。
class Solution(object): def titleToNumber(self, s): """ :type s: str :rtype: int """ letters = {"A": 1, "B": 2, "C": 3, "D": 4, "E": 5, "F": 6, "G": 7, "H": 8, "I": 9, "J": 10, "K": 11, "L": 12, "M": 13, "N": 14, "O": 15, "P": 16, "Q": 17, "R": 18, "S": 19, "T":20, "U": 21, "V": 22, "W": 23, "X": 24, "Y": 25, "Z": 26} s = s.upper() ret = 0 b = 0 for c in reversed(s): ret += letters[c] * 26**(b) b += 1 return ret我的想法和大佬的大致还是差不多的哈，开心
