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

- 日理万妓 2023-09-23 21:28 46阅读 0赞

# 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0
            if(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0
            int 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++;
                else
                    res--;
            }
            //3.返回res值
            return res;     
        }
    }