从两个数字的总和中删除尾随零（使用堆栈）

给定两个数字A和B ，任务是使用堆栈删除两个给定数字之和中存在的尾随零。

例子：

Input: A = 124, B = 186
Output: 31
Explanation: Sum of A and B is 310. Removing the trailing zeros modifies the sum to 31.

Input: A=130246, B= 450164
Output : 58041

方法：可以使用字符串和堆栈数据结构解决给定的问题。 请按照以下步骤解决问题：

• 计算A + B并将其存储在一个变量中，例如N 。
• 初始化一个堆栈 ，比如说S ，来存储N的数字。
• 将整数N转换为字符串，然后将字符压入堆栈S 。
• 迭代while S不为空()，如果栈顶元素为’0’，则将其从栈中弹出。否则，断。
• 初始化一个字符串，比如res ，以存储结果字符串。
• 迭代而 S 不为空()，将res 中的所有字符压入，然后弹出顶部元素。
• 反转字符串res并打印res作为答案。

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to remove trailing
// zeros from the sum of two numbers
string removeTailing(int A, int B)
{
    // Stores the sum of A and B
    int N = A + B;
 
    // Stores the digits
    stack s;
 
    // Stores the equivalent
    // string of integer N
    string strsum = to_string(N);
 
    // Traverse the string
    for (int i = 0; i < strsum.length(); i++) {
 
        // Push the digit at i
        // in the stack
        s.push(strsum[i]);
    }
 
    // While top element is '0'
    while (s.top() == '0')
 
        // Pop the top element
        s.pop();
 
    // Stores the resultant number
    // without tailing 0's
    string res = "";
 
    // While s is not empty
    while (!s.empty()) {
 
        // Append top element of S in res
        res = res + char(s.top());
 
        // Pop the top element of S
        s.pop();
    }
 
    // Reverse the string res
    reverse(res.begin(), res.end());
 
    return res;
}
 
// Driver Code
int main()
{
    // Input
    int A = 130246, B = 450164;
 
    // Function Call
    cout << removeTailing(A, B);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to remove trailing
// zeros from the sum of two numbers
static String removeTailing(int A, int B)
{
     
    // Stores the sum of A and B
    int N = A + B;
 
    // Stores the digits
    Stack s = new Stack();
 
    // Stores the equivalent
    // string of integer N
    String strsum = Integer.toString(N);
 
    // Traverse the string
    for(int i = 0; i < strsum.length(); i++)
    {
         
        // Push the digit at i
        // in the stack
        s.push(strsum.charAt(i));
    }
 
    // While top element is '0'
    while (s.peek() == '0')
    {
         
        // Pop the top element
        s.pop();
    }
 
    // Stores the resultant number
    // without tailing 0's
    String res = "";
 
    // While s is not empty
    while (s.empty() == false)
    {
         
        // Append top element of S in res
        res = res + (char)s.peek();
 
        // Pop the top element of S
        s.pop();
    }
 
    StringBuilder str = new StringBuilder();
    str.append(res);
 
    // Reverse the string res
    str.reverse();
 
    return str.toString();
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Input
    int A = 130246, B = 450164;
 
    // Function Call
    System.out.println(removeTailing(A, B));
}
}
 
// This code is contributed by Dharanendra.L.V.

Python3

# Python 3 program for the above approach
 
# Function to remove trailing
# zeros from the sum of two numbers
def removeTailing(A,  B):
 
    # Stores the sum of A and B
    N = A + B
 
    # Stores the digits
    s = []
 
    # Stores the equivalent
    # string of integer N
    strsum = str(N)
 
    # Traverse the string
    for i in range(len(strsum)):
 
        # Push the digit at i
        # in the stack
        s.append(strsum[i])
 
    # While top element is '0'
    while (s[-1] == '0'):
 
        # Pop the top element
        s.pop()
 
    # Stores the resultant number
    # without tailing 0's
    res = ""
 
    # While s is not empty
    while (len(s) != 0):
 
        # Append top element of S in res
        res = res + (s[-1])
 
        # Pop the top element of S
        s.pop()
 
    # Reverse the string res
    res = list(res)
    res.reverse()
    res = ''.join(res)
 
    return res
 
 
# Driver Code
if __name__ == "__main__":
 
    # Input
    A = 130246
    B = 450164
 
    # Function Call
    print(removeTailing(A, B))
 
    # This code is contributed by ukasp.

Javascript

 
        // Javascript program for the above approach
 
        // Function to remove trailing
        // zeros from the sum of two numbers
        function removeTailing(A, B) {
 
            // Stores the sum of A and B
            let N = A + B;
 
            // Stores the digits
            let s = new Array();
 
            // Stores the equivalent
            // string of integer N
            let strsum = N.toString();
 
            // Traverse the string
            for (let i = 0; i < strsum.length; i++) {
 
                // Push the digit at i
                // in the stack
                s.push(strsum.charAt(i));
            }
 
            // While top element is '0'
            while (s[s.length-1] === '0') {
 
                // Pop the top element
                s.pop();
            }
        
            // Stores the resultant number
            // without tailing 0's
            let res = "";
 
            // While s is not empty
            while (s.length != 0) {
 
                // Append top element of S in res
                res = res.concat(s[s.length-1]);
 
                // Pop the top element of S
                s.pop();
            }
 
            let str = "";
            str = str.concat(res)
 
            // Reverse the string res
            str = str.split("").reverse().join("");
 
            return str.toString();
        }
 
        // Driver Code
 
        // Input
        let A = 130246, B = 450164;
 
        // Function Call
        document.write(removeTailing(A, B));
 
        // This code is contributed by Hritik
         
    

输出：

58041

时间复杂度： O(len(A + B))
辅助空间： O(len(A + B))