将以下内容翻译为中文：尽量减少移除非相邻相同字符的次数，以使给定的字符串变为空字符串

在本文中，我们将深入探讨C++中一个引人入胜的字符串操作问题。问题陈述是“最小化删除非相邻字符以使给定字符串为空”。这个问题是提升你对字符串、字符删除和算法思维的理解的绝佳方式。

Problem Statement

Given a string, the task is to minimize the number of removal operations of non-equal adjacent characters required to make the given string empty. In one operation, you can remove any two adjacent characters that are not equal.

C++ Solution Approach

解决这个问题的方法是使用堆栈数据结构。我们遍历字符串的字符，对于每个字符，如果堆栈不为空且堆栈的顶部字符不等于当前字符，则从堆栈中弹出顶部字符。否则，将当前字符推入堆栈。所需的操作次数是最后堆栈中剩余字符的数量。

Example

#include <iostream>
#include <stack>
#include <string>
using namespace std;

int minimizeRemovals(string str) {
   stack<char> s;
   for (char c : str) {
      if (!s.empty() && s.top() != c) {
         s.pop();
      } else {
         s.push(c);
      }
   }
   return s.size();
}

int main() {
   string str = "abba";
   int operations = minimizeRemovals(str);
   cout << "The minimum number of removal operations is: " << operations << endl;
   return 0;
}

Output

The minimum number of removal operations is: 0

Explanation with a Test Case

当我们将这个字符串传递给minimizeRemovals函数时，它会迭代字符串的字符。过程如下所示−

• 它将 'a' 推入堆栈。

• 然后它将 'b' 推入堆栈，因为 'b' 不等于堆栈顶部的元素 ('a')。

• When the next 'b' is encountered, it sees that the top of the stack is also 'b', so it doesn't perform a remove operation, and 'b' is pushed onto the stack.

• Now the top of the stack is 'b', and the next character is 'a'. Since 'a' is not equal to 'b', it performs a remove operation by popping the top of the stack. Now the top of the stack is 'b'.

• 最后，在字符串中遇到了与栈顶元素（'b'）不相等的字符'a'。因此，它执行了一个移除操作，弹出了栈顶元素。

At the end of the function, there are no characters left in the stack, indicating that all non-equal adjacent characters have been removed from the string. Hence, the function returns 0, which is the minimum number of removal operations required to make the given string empty.

Conclusion

这个问题为使用堆栈数据结构进行字符串操作提供了绝佳的机会。这是一个很好的问题，可以练习你的C++编码技巧，并理解如何使用堆栈解决问题。

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