408 King's Road Data Structure After-School Exercises (1)

Get into the habit of writing together! This is the 5th day of my participation in the "Nuggets Daily New Plan · April Update Challenge", click to view the details of the event .

Open a new pit, and plan to update the implementation of all after-school code exercises in the 23 Wangdao data structure.

Although the exams are generally written in pseudocode, I have implemented them all because of obsessive-compulsive disorder. The warehouse is here

sequence table structure

The test is used directly, generally will not let you write the structure

#define MaxSize 50

typedef struct {
  ElemType data[MaxSize];
  int length;
}SqList; 
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2.2.3, 1

• The violent solution is generally a cycle. If the cycle fails once, it will come twice, and if it fails twice, it will be repeated three times...
• Time complexity O(n), space complexity O(1)

int del_min(SqList &list) {
  if (list.length == 0) {
    cout << "Error!" << endl;
    return -1;
  }

  // 1.假设0号元素最小
  int min = list.data[0]; 
  int pos = 0;

  // 2.循环找出最小元素并记录位置, 从1开始
  for (int i = 1; i < list.length; i++) {
    if (list.data[i] < min) {
      min = list.data[i];
      pos = i;
    }
  }

  // 3.删除最小元素并用最后一个元素替换
  list.data[pos] = list.data[list.length - 1];
  list.length--; 

  return min;
}
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2.2.3, 2

• Violence is another whole array, the original array is traversed from the end and placed into the new array
• If the space complexity is O(1), loop from the beginning to the middle, and exchange the head and tail elements
• Even or odd number of elements, loop to the middle, all < length/2, because if the length is 4 or 5, the loop stops when the subscript is 1, and when it is 5, the middle element (subscript 2) does not need to move.
• Time complexity O(n)

void reverse(SqList &list) {
  for (int i = 0; i < list.length / 2; i++) {
    // 不让用swap()的话，可以定义一个辅助变量来交换
    swap(list.data[i], list.data[list.length - 1 - i]);
  }
}
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2.2.3, 3

• Violence is to complete another array, loop through and put all elements not equal to x into the new array
• Time complexity O(n), space complexity O(1) requires one loop to solve
• Throw all elements equal to x to the end, and then subtract the length
• Reverse: that is, throw all elements that are not equal to x to the front, and naturally the elements that are equal to x are behind.

void del_x(SqList &list, int x) {
  int k = 0;
  for (int i = 0 ; i < list.length ; i++) {
    // 1.把所有要保存的值都放在前面
    if (list.data[i] != x) {
      list.data[k++] = list.data[i];
    }
  }
  // 2.直接扔掉后面的元素
  list.length = k;
}
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• One loop -> double pointer
• Similar to quicksort, moving from both ends to the middle, swapping the x on the left with the non-x on the right

// 太麻烦了，不如上一个方法
void del_x_2(SqList &list, int x) {
  int i = -1, j = list.length, k = 0;
  while (i < j) {
    while (list.data[++i] != x); 
    while (list.data[--j] == x) k++;
    if (i < j) {
      swap(list.data[i], list.data[j]);
      k++;
    }
  }
  list.length -= k;
}
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