给定一个由N个字符串组成的数组arr [] ，任务是查找遍历给定数组arr []时执行以下操作构成的brr []数组的总和(最初为空)：
- 如果数组arr []包含一个整数，则将该整数插入到数组brr []中。
- 如果数组arr []的字符串“ +” ，则将数组brr []中最后两个元素的和插入数组brr []中。
- 如果数组arr []的字符串“ D” ，则将值brr []的最后一个元素的两倍插入数组brr [] 。
- 如果数组arr []的字符串“ C” ，则将数组brr []的最后一个元素删除到数组brr []中。
例子：
Input: arr[] = {“5”, “2”, “C”, “D”, “+”}
Output: 30
Explanation:
While traversing the array arr[], the array brr[] is modified as:
- “5” – Add 5 to the array brr[]. Now, the array brr[] modifies to {5}.
- “2” – Add 2 to the array brr[]. Now, the array brr[] modifies to {5, 2}.
- “C” – Remove the last element from the array brr[]. Now, the array brr[] modifies to {5}.
- “D” – Add twice the last element of the array brr[] to the array brr[]. Now, the array brr[] modifies to {5, 10}.
- “+” – Add the sum of the last two elements of the array brr[] to the array brr[]. Now the array brr[] modifies to {5, 10, 15}.
After performing the above operations, the total sum of the array brr[] is 5 + 10 + 15 = 30.
Input: arr[] = {“5”, “-2”, “4”, “C”, “D”, “9”, “+”, “+”}
Output: 27
方法：解决给定问题的想法是使用堆栈。请按照以下步骤解决问题：
- 初始化一个整数堆栈(例如S) ，并初始化一个变量(例如ans)为0 ，以存储形成的数组的结果总和。
- 遍历给定数组arr []并执行以下步骤：
- 如果arr [i]的值为“ C” ，则从ans中减去堆栈的顶部元素，然后从S中弹出它。
- 如果arr [i]的值为“ D” ，则在堆栈S中将堆栈S的顶部元素压入两次，然后将其值添加到ans 。
- 如果arr [i]的值为“ +” ，则推入堆栈S的前两个元素之和的值，并将它们的和加到ans上。
- 否则，将arr [i]推入堆栈S ，并将其值添加到ans 。
- 循环之后，将ans的值打印为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the sum of the array
// formed by performing given set of
// operations while traversing the array ops[]
void findTotalSum(vector& ops)
{
// If the size of array is 0
if (ops.empty()) {
cout << 0;
return;
}
stack pts;
// Stores the required sum
int ans = 0;
// Traverse the array ops[]
for (int i = 0; i < ops.size(); i++) {
// If the character is C, remove
// the top element from the stack
if (ops[i] == "C") {
ans -= pts.top();
pts.pop();
}
// If the character is D, then push
// 2 * top element into stack
else if (ops[i] == "D") {
pts.push(pts.top() * 2);
ans += pts.top();
}
// If the character is +, add sum
// of top two elements from the stack
else if (ops[i] == "+") {
int a = pts.top();
pts.pop();
int b = pts.top();
pts.push(a);
ans += (a + b);
pts.push(a + b);
}
// Otherwise, push x
// and add it to ans
else {
int n = stoi(ops[i]);
ans += n;
pts.push(n);
}
}
// Print the resultant sum
cout << ans;
}
// Driver Code
int main()
{
vector arr = { "5", "-2", "C", "D", "+" };
findTotalSum(arr);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
// Function to find the sum of the array
// formed by performing given set of
// operations while traversing the array ops[]
static void findTotalSum(String ops[])
{
// If the size of array is 0
if (ops.length == 0)
{
System.out.println(0);
return;
}
Stack pts = new Stack<>();
// Stores the required sum
int ans = 0;
// Traverse the array ops[]
for (int i = 0; i < ops.length; i++) {
// If the character is C, remove
// the top element from the stack
if (ops[i] == "C") {
ans -= pts.pop();
}
// If the character is D, then push
// 2 * top element into stack
else if (ops[i] == "D") {
pts.push(pts.peek() * 2);
ans += pts.peek();
}
// If the character is +, add sum
// of top two elements from the stack
else if (ops[i] == "+") {
int a = pts.pop();
int b = pts.peek();
pts.push(a);
ans += (a + b);
pts.push(a + b);
}
// Otherwise, push x
// and add it to ans
else {
int n = Integer.parseInt(ops[i]);
ans += n;
pts.push(n);
}
}
// Print the resultant sum
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
String arr[] = { "5", "-2", "C", "D", "+" };
findTotalSum(arr);
}
}
// This code is contributed by Kingash. Python3
# Python3 program for the above approach
# Function to find the sum of the array
# formed by performing given set of
# operations while traversing the array ops[]
def findTotalSum(ops):
# If the size of array is 0
if (len(ops) == 0):
print(0)
return
pts = []
# Stores the required sum
ans = 0
# Traverse the array ops[]
for i in range(len(ops)):
# If the character is C, remove
# the top element from the stack
if (ops[i] == "C"):
ans -= pts[-1]
pts.pop()
# If the character is D, then push
# 2 * top element into stack
elif (ops[i] == "D"):
pts.append(pts[-1] * 2)
ans += pts[-1]
# If the character is +, add sum
# of top two elements from the stack
elif (ops[i] == "+"):
a = pts[-1]
pts.pop()
b = pts[-1]
pts.append(a)
ans += (a + b)
pts.append(a + b)
# Otherwise, push x
# and add it to ans
else:
n = int(ops[i])
ans += n
pts.append(n)
# Print the resultant sum
print(ans)
# Driver Code
if __name__ == "__main__":
arr = ["5", "-2", "C", "D", "+"]
findTotalSum(arr)
# This code is contributed by ukasp.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the sum of the array
// formed by performing given set of
// operations while traversing the array ops[]
static void findTotalSum(string []ops)
{
// If the size of array is 0
if (ops.Length == 0)
{
Console.WriteLine(0);
return;
}
Stack pts = new Stack();
// Stores the required sum
int ans = 0;
// Traverse the array ops[]
for(int i = 0; i < ops.Length; i++)
{
// If the character is C, remove
// the top element from the stack
if (ops[i] == "C")
{
ans -= pts.Pop();
}
// If the character is D, then push
// 2 * top element into stack
else if (ops[i] == "D")
{
pts.Push(pts.Peek() * 2);
ans += pts.Peek();
}
// If the character is +, add sum
// of top two elements from the stack
else if (ops[i] == "+")
{
int a = pts.Pop();
int b = pts.Peek();
pts.Push(a);
ans += (a + b);
pts.Push(a + b);
}
// Otherwise, push x
// and add it to ans
else
{
int n = Int32.Parse(ops[i]);
ans += n;
pts.Push(n);
}
}
// Print the resultant sum
Console.WriteLine(ans);
}
// Driver Code
public static void Main()
{
string []arr = { "5", "-2", "C", "D", "+" };
findTotalSum(arr);
}
}
// This code is contributed by ipg2016107 输出：
30时间复杂度： O(N)
辅助空间： O(N)