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 )