执行操作后的变量值【LC2011】

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.

再去练几道状态压缩dp！

• 思路：遍历整个数组，模拟加一和减一的操作，最后返回结果。进行加一或者减一的操作取决于字符串的第二个字符：

。如果第二个字符是+号，那么执行加一操作

。如果第二个字符是-号，那么执行减一操作

• 实现

class Solution {
    public int finalValueAfterOperations(String[] operations) {
        int res = 0;
        for (String operation : operations){
            res += operation.charAt(1) == '+' ? 1 : -1;
        }
        return res;
    }
}

。复杂度

• 时间复杂度：O(n),n为数组长度
• 空间复杂度：O ( 1 )