[Leet Code - eazy] 2011. Final Value of Variable After Performing Operations

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.

 

나의 풀이

 

/**
 * @param {string[]} operations
 * @return {number}
 */
var finalValueAfterOperations = function(operations) {
    let answer = 0
    for(let i of operations){
        switch(i){
            case '++X':
            case 'X++':
            answer+=1
            break
 
            case '--X':
            case 'X--':
            answer-=1
            break
        }
       
    }
    return answer 
};

 

문제를 풀며 느낀 점

- for of를 통해서 배열의 value를 하나하나 i로 순회하여 i를 switch문의 조건에 따라서 처리하였습니다.