1.问题

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.

2.解决思路

方法1：

1.判断数据是存在，不存在返回0
2.设置一个int[]temp数组，长度和String[] operations的长度相同
3.将String[] operations的元素遍历，进行判断并赋值为整数的值放入temp数组中
4.遍历int[]temp的元素进行求和

方法2：

1.定义一个变量res，初始值为0
2.遍历String[] operations，用contains() Method如果元素包含“+”，在res上加一，否则减-
3.返回res值

3.代码实现

代码1：

class Solution {public int finalValueAfterOperations(String[] operations) {//1.判断数据是存在，不存在返回0if(operations == null){return 0;}//2.设置一个int[]temp数组，长度和String[] operations的长度相同int[] temp = new int[operations.length]; //3.将String[] operations的元素遍历，进行判断并赋值为整数的值放入temp数组中for(int i=0;i<operations.length;i++){if(operations[i].equals("X++") || operations[i].equals("++X")){temp[i] = 1;}else{temp[i] = -1;}}//4.遍历int[]temp的元素进行求和int sum=0;for(int j=0;j<temp.length;j++){sum += temp[j];}return sum;       }
}

代码2

class Solution {public int finalValueAfterOperations(String[] operations) {//1.定义一个变量res，初始值为0int res = 0;//2.遍历String[] operations，用contains() Method如果元素包含“+”，在res上加一，否则减-for(int i = 0; i < operations.length; i++){if(operations[i].contains("+"))res++;elseres--;}//3.返回res值return res;     }
}

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LeetCode（String) 2011. Final Value of Variable After Performing Operations

1.问题

2.解决思路

3.代码实现

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