描述

The letter value of a letter is its position in the alphabet starting from 0 (i.e. ‘a’ -> 0, ‘b’ -> 1, ‘c’ -> 2, etc.).

The numerical value of some string of lowercase English letters s is the concatenation of the letter values of each letter in s, which is then converted into an integer.

For example, if s = “acb”, we concatenate each letter’s letter value, resulting in “021”. After converting it, we get 21.
You are given three strings firstWord, secondWord, and targetWord, each consisting of lowercase English letters ‘a’ through ‘j’ inclusive.

Return true if the summation of the numerical values of firstWord and secondWord equals the numerical value of targetWord, or false otherwise.

Example 1:

Input: firstWord = "acb", secondWord = "cba", targetWord = "cdb"
Output: true
Explanation:
The numerical value of firstWord is "acb" -> "021" -> 21.
The numerical value of secondWord is "cba" -> "210" -> 210.
The numerical value of targetWord is "cdb" -> "231" -> 231.
We return true because 21 + 210 == 231.

Example 2:

Input: firstWord = "aaa", secondWord = "a", targetWord = "aab"
Output: false
Explanation: 
The numerical value of firstWord is "aaa" -> "000" -> 0.
The numerical value of secondWord is "a" -> "0" -> 0.
The numerical value of targetWord is "aab" -> "001" -> 1.
We return false because 0 + 0 != 1.

Example 3:

Input: firstWord = "aaa", secondWord = "a", targetWord = "aaaa"
Output: true
Explanation: 
The numerical value of firstWord is "aaa" -> "000" -> 0.
The numerical value of secondWord is "a" -> "0" -> 0.
The numerical value of targetWord is "aaaa" -> "0000" -> 0.
We return true because 0 + 0 == 0.

Note:

1 <= firstWord.length, secondWord.length, targetWord.length <= 8
firstWord, secondWord, and targetWord consist of lowercase English letters from 'a' to 'j' inclusive.

解析

根据题意，就是将 firstWord 、 secondWord 和 targetWord 的字符串转换成对应的数字，判断前面两者的和是否和后者相等。思路简单，就是定一个将字符串转换成数字的函数，关键注意判断全 0 字符串转换为整数位 0 ，其他的需要调用函数进行布尔运算即可。

解答

class Solution(object):
    def isSumEqual(self, firstWord, secondWord, targetWord):
        """
        :type firstWord: str
        :type secondWord: str
        :type targetWord: str
        :rtype: bool
        """
        def convert(s):
            result = ''
            for c in s:
                result += str(ord(c)-97)
            if result.count('0') == len(result):
                return 0
            return int(result.lstrip('0'))
        return convert(firstWord) + convert(secondWord) == convert(targetWord)

运行结果

Runtime: 12 ms, faster than 96.48% of Python online submissions for Check if Word Equals Summation of Two Words.
Memory Usage: 13.4 MB, less than 59.93% of Python online submissions for Check if Word Equals Summation of Two Words.

原题链接：https://leetcode.com/problems/check-if-word-equals-summation-of-two-words

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