What does the 'Score of a String' problem measure?

It measures the sum of absolute differences between ASCII values of adjacent characters in the string.

Can this solution handle very long strings?

Yes, the approach is O(n) in time and O(1) in space, efficiently handling strings up to length 100.

Do I need to store differences in an array?

No, you can maintain a running total of differences, keeping memory usage constant.

How do repeated characters affect the score?

Repeated characters contribute zero to the score since their ASCII difference is zero.

What is the key pattern for solving this problem?

The key pattern is a string-driven solution strategy where you iterate through characters and sum absolute differences efficiently.

#3110

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

字符串的分数

给你一个字符串 s 。一个字符串的分数定义为相邻字符 ASCII 码差值绝对值的和。请你返回 s 的分数。示例 1：输入： s = "hello" 输出： 13 解释： s 中字符的 ASCII 码分别为： 'h' = 104 ， 'e' = 101 ， 'l' = 108 ， 'o'…

字符串

题目描述

给你一个字符串 s 。一个字符串的分数定义为相邻字符 ASCII 码差值绝对值的和。

请你返回 s 的分数。

示例 1：

输入：s = "hello"

输出：13

解释：

s 中字符的 ASCII 码分别为：'h' = 104 ，'e' = 101 ，'l' = 108 ，'o' = 111 。所以 s 的分数为 |104 - 101| + |101 - 108| + |108 - 108| + |108 - 111| = 3 + 7 + 0 + 3 = 13 。

示例 2：

输入：s = "zaz"

输出：50

解释：

s 中字符的 ASCII 码分别为：'z' = 122 ，'a' = 97 。所以 s 的分数为 |122 - 97| + |97 - 122| = 25 + 25 = 50 。

提示：

2 <= s.length <= 100
s 只包含小写英文字母。

lightbulb

解题思路

方法一：模拟

我们直接遍历字符串 $s$ ，计算相邻字符的 ASCII 码差值的绝对值之和即可。

时间复杂度 $O(n)$ ，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def scoreOfString(self, s: str) -> int:
        return sum(abs(a - b) for a, b in pairwise(map(ord, s)))

speed

复杂度分析

指标	值
时间	complexity is O(n) because each character is visited once to compute its difference with the next. Space complexity is O(1) since only a running total and the previous character need storage.
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Check if candidates correctly use ASCII values for adjacent characters.
question_mark
Watch for unnecessary data structures that increase space usage.
question_mark
Listen for clear explanation of the string-driven iterative pattern.

warning

常见陷阱

外企场景

error
Forgetting to take the absolute value of differences.
error
Attempting to store all differences instead of a running sum.
error
Failing on repeated characters where difference is zero.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Compute the score using the difference between Unicode code points for extended character sets.
arrow_right_alt
Return a cumulative array of scores at each position instead of a total sum.
arrow_right_alt
Apply the same scoring method to a list of strings and return their maximum score.

help

常见问题

外企场景

继续练习

#3106 满足距离约束且字典序最小的字符串 #3114 替换字符可以得到的最晚时间 #3120 统计特殊字母的数量 I