[리트코드LeetCode][EASY][JAVA]2011. Final Value of Variable After Performing Operations

 

문제

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--".

 

 

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분석

1. ++이나 -- 가 앞에 붙든 뒤에 붙든 관계 없이 기호에 따라 return 할 값(x)을 +1 또는 -1 해서 최종적으로 return하면 됨.

 

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답안

class Solution {
    public int finalValueAfterOperations(String[] operations) {
        int answer = 0;

        for(String s : operations){
            char c = s.charAt(1);
            switch(c){
                case '+' : 
                    answer++;
                    break;
                case '-' :
                    answer--;
                    break;
            }
        }
        return answer;
    }
}

 

 

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다른 사람의 답안

// 메모리 효율이 최상위인 코드

class Solution {
    public int finalValueAfterOperations(String[] operations) {
        int count=0;
        for(int i=0;i<operations.length;i++){
            if(operations[i].contains("+"))
                count++;
            else
                count--;
        }
        return count;
        
    }
}

 

 

 

 

 

 

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