408 After-school code exercises for kingly data structure (twenty-three)

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foreword

It is planned to update the implementation of all after-school code exercises in the 23 King's Road data structure. Although the exams are generally written in pseudo-code, I have implemented them all because of obsessive-compulsive disorder. The warehouse is here.

The code is written in C++, all of which can be compiled and run, including violent solutions and optimal solutions as much as possible (if the exam time is not enough, you can directly attack a violent one, and you can get more than half of the points).

Continuous update, current update progress:

for reference only! There will be some grammar that the exam is not allowed to write, and there may be some mistakes.

3.1.5, 3

The initial state and final state of the stack clearly written in the question are empty, so the number of times of stacking and popping should be the same
The stack is recorded as +1, the pop-out is -1, and the final result is 0, which is legal. If there is a case of <0 in the middle process, it is illegal.
Time complexity O(n), space complexity O(1)

bool isValid(string str) {
  int cnt = 0;
  for (int i = 0; i < str.length(); i++) {
    if (str[i] == 'I') cnt++;
    if (str[i] == 'O') cnt--;

    if (cnt < 0) return false;
  }

  if (cnt == 0) return true;
  return false;
}
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4

Centrosymmetric
Traverse, use the stack to store the first half of the elements, and then compare the second half of the elements
Only need to pay attention to the determination of odd and even elements
Time complexity O(n), space complexity O(n/2)

bool isSymmetry(LinkList L, int n) {
  char stk[n/2];      // 字符栈
  LNode *p = L->next;

  // 1.存入链表前一半的元素
  for (int i = 0; i < n/2; i++) {
    stk[i] = p->data;
    p = p->next;
  }

  // 2.如果链表元素个数为奇数，则中间元素不用比较
  if (n % 2 == 1) p = p->next;

  // 3.对比后半部分
  for (int i = n/2 - 1; i >= 0; i--) {
    if (stk[i] != p->data) return false;
    p = p->next;
  }
  return true;
}
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5

The key to sharing the sequential stack is the determination of the two stations into and out of the stack. s1 is a stack in the traditional sense. When the stack is pushed, the top pointer of the stack is +1, and s2 is the opposite. When the stack is pushed, the top pointer of the stack is -1. pop the stack
Note that "the stack is full, and the stack is empty "
The way to judge the stack is full istop2-top1 == 1

#define maxsize 100
typedef struct {
  int stack[maxsize];
  int top1;
  int top2;
} stk;

// 栈的初始化
void initStack(stk &s) {
  s.top1 = -1;
  s.top2 = maxsize;
}

// 判断栈空
bool isEmpty(stk &s, int num) {
  if (num != 1 && num != 2) {
    cout << "栈号输入错误" << endl;
    return false;
  }

  if (num == 1) return s.top1 == -1;
  return s.top2 == maxsize;
}

// 判断栈满
bool isFull(stk &s) {
  if (s.top2 - s.top1 == 1) return true;
  return false;
}

// 入栈
bool push(stk &s, int num, int x) {
  if (num != 1 && num != 2) {
    cout << "栈号输入错误" << endl;
    return false;
  }

  if (isFull(s)) {
    cout << "栈满" << endl;
    return false;
  }

  if (num == 1) s.stack[++s.top1] = x;
  else s.stack[--s.top2] = x;
  return true;
}

// 出栈
bool pop(stk &s, int num, int &x) {
  if (num != 1 && num != 2) {
    cout << "栈号输入错误" << endl;
    return false;
  }

  if (isEmpty(s, num)) {
    cout << "栈空" << endl;
    return false;
  }

  if (num == 1) x = s.stack[s.top1--];
  else x = s.stack[s.top2++];
  return true;
}
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