2011.final-value-of-variable-after-performing-operations

Statement

简体中文English

Metadata

• Link: 执行操作后的变量值
• Difficulty: Easy
• Tag: 数组 字符串 模拟

存在一种仅支持 4 种操作和 1 个变量 X 的编程语言：

• ++X 和 X++ 使变量 X 的值 加 1
• –X 和 X– 使变量 X 的值 减 1

最初，X 的值是 0

给你一个字符串数组 operations ，这是由操作组成的一个列表，返回执行所有操作后， X 的 最终值 。

 

示例 1：

输入：operations = ["–X","X++","X++"]
输出：1
解释：操作按下述步骤执行：
最初，X = 0
–X：X 减 1 ，X =  0 - 1 = -1
X++：X 加 1 ，X = -1 + 1 =  0
X++：X 加 1 ，X =  0 + 1 =  1

示例 2：

输入：operations = ["++X","++X","X++"]
输出：3
解释：操作按下述步骤执行： 
最初，X = 0
++X：X 加 1 ，X = 0 + 1 = 1
++X：X 加 1 ，X = 1 + 1 = 2
X++：X 加 1 ，X = 2 + 1 = 3

示例 3：

输入：operations = ["X,X","–X","X–"]
输出：0
解释：操作按下述步骤执行：
最初，X = 0
X++：X 加 1 ，X = 0 + 1 = 1
++X：X 加 1 ，X = 1 + 1 = 2
–X：X 减 1 ，X = 2 - 1 = 1
X–：X 减 1 ，X = 1 - 1 = 0

 

提示：

• 1 <= operations.length <= 100
• operations[i] 将会是 "++X"、"X++"、"–X" 或 "X–"

Metadata

• Link: Final Value of Variable After Performing Operations
• Difficulty: Easy
• Tag: Array String Simulation

There is a programming language with only four operations and one variable X:

• ++X and X++ increments the value of the variable X by 1.
• –X and X– decrements the value of the variable X by 1.

Initially, the value of X is 0.

Given an array of strings operations containing a list of operations, return the final value of X after performing all the operations.

 

Example 1:

Input: operations = ["–X","X++","X++"]
Output: 1
Explanation: The operations are performed as follows:
Initially, X = 0.
–X: X is decremented by 1, X =  0 - 1 = -1.
X++: X is incremented by 1, X = -1 + 1 =  0.
X++: X is incremented by 1, X =  0 + 1 =  1.

Example 2:

Input: operations = ["++X","++X","X++"]
Output: 3
Explanation: The operations are performed as follows:
Initially, X = 0.
++X: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
X++: X is incremented by 1, X = 2 + 1 = 3.

Example 3:

Input: operations = ["X++","++X","–X","X–"]
Output: 0
Explanation: The operations are performed as follows:
Initially, X = 0.
X++: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
–X: X is decremented by 1, X = 2 - 1 = 1.
X–: X is decremented by 1, X = 1 - 1 = 0.

 

Constraints:

• 1 <= operations.length <= 100
• operations[i] will be either "++X", "X++", "–X", or "X–".

Solution

Python 3

from typing import List


class Solution:
    def finalValueAfterOperations(self, operations: List[str]) -> int:
        def f(s):
            if "+" in s:
                return 1
            return -1
        return sum(map(f, operations))

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最后更新: October 11, 2023