2011.md

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2011. 执行操作后的变量值	LeetCode 2011. 执行操作后的变量值题解，Final Value of Variable After Performing Operations，包含解题思路、复杂度分析以及完整的 JavaScript 代码实现。	LeetCode 2011. 执行操作后的变量值 执行操作后的变量值 Final Value of Variable After Performing Operations 解题思路 数组 字符串 模拟

2011. 执行操作后的变量值

🟢 Easy  🔖  数组 字符串 模拟  🔗 力扣 LeetCode

题目

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 _thefinal 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--".

题目大意

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--"

解题思路

1. 初始化变量 x = 0。

2. 遍历 operations 数组，检查每个字符串的中心字符。

   • 操作 "++X" 和 "X++" 的中心字符是 '+'(ASCII 值 43)，表示增加 1。
   • 操作 "--X" 和 "X--" 的中心字符是 '-'(ASCII 值 45)，表示减少 1。

3. 通过 charCodeAt(1) 提取中心字符的 ASCII 值，再通过44 - str.charCodeAt(1) 计算操作的数值：

   • 如果中心字符为 '+'：44 - 43 = 1。
   • 如果中心字符为 '-'：44 - 45 = -1。

4. 累加所有的变化到 x，返回结果。

复杂度分析

• 时间复杂度：O(n)，其中 n 是 operations 的长度，遍历数组一次。
• 空间复杂度：O(1)，使用常数额外空间变量。

代码

/**
 * @param {string[]} operations
 * @return {number}
 */
var finalValueAfterOperations = function (operations) {
	let x = 0;
	operations.forEach((str) => (x += 44 - str.charCodeAt(1)));
	return x;
};