【Algorithm Training: Linear Tables】Day 06

topic

Assuming that n (n>1) integers are stored in a one-dimensional array R, design an algorithm that is as efficient as possible in terms of time and space. Move the sequence saved in R to the left by p (0<p<n) positions, that is, transform the data in R from (x0,x1,x2,…,xn-1) to (xp,xp+1,…, xn+1,x0,x1,...,xp-1), requires:

1. The basic design idea of ​​the algorithm is given.
2. According to the design idea, the algorithm is implemented in C/C++, and comments are given at the key points.
3. Explain the time complexity and space complexity of the algorithm.

Basic Design Ideas

1. Based on the above requirements, it is analyzed that the title mainly requires that the elements in the array R be cyclically shifted to the left by p bits, and return to the realized array.
2. Therefore, in the first step, we can first consider reversing all the elements in the array R to obtain the array (xn-1, xn-2,..., x2, x1, x0).
3. Next, reverse the first np elements in the array R to obtain the array (xp,xp+1,...,xn-1,xp-1,xp-2,...x2,x1,x0).
4. Then, reverse the last p elements in the array R to get the array (xp, xp+1,...,xn-1, x0, x1,...,xp-1) required by the title.

Code

Array LinearList::Reverse(int left,int right)//传入顺序表arr逆置的左右临界
{
    
    
    if(arr.length <= 1)
        return arr;

    int i = left,j = right-1;//设置哨兵i，j分别指向左右临界的下标
    int temp = 0;
    while(i < j)    //交换哨兵指向的元素位置
    {
    
    
        temp = arr.data[i];
        arr.data[i++] = arr.data[j];
        arr.data[j--] = temp;
        temp = 0;
    }
    return arr;
}

Array LinearList::Question_10(int p)
{
    
    
    if(arr.length <=1)//判断非法数组长度
        return arr;
    cout<<"Move "<<p<<" bit left"<<endl;
    Reverse(0,arr.length); //逆置数组R中的全部元素
    Reverse(0,arr.length-p);   //逆置数组R中前n-p个元素
    Reverse(arr.length-p,arr.length);  //逆置数组R中后p个元素

    return arr;
}

Effect