创新工场2014笔试算法题汇总

2024年3月30日 114次阅读

1. 堆排序

#include<iostream>   
usingnamespace std;   
  
void SwapValue(int &m, int &n)  
{  
    int temp = m;  
    m = n;  
    n = temp;  
}  
void max_heap(vector<int> &vec, int i, int heap_size)  
{  
    int l = 2*i;  
    int r = 2*i+1;  
    int largest = i;  
      
    if(l<=heap_size && vec[l-1]>vec[largest-1])  
        largest = l;  
    if(r<=heap_size && vec[r-1]>vec[largest-1])  
        largest = r;  
  
    if(largest!=i)  
    {  
        SwapValue(vec[largest-1],vec[i-1]);  
        max_heap(vec, largest, heap_size);  
    }  
}  
void heapSort(vector<int> &vec)  
{  
    int heap_size = vec.size();  
    for(int i=heap_size/2; i>=1; i--)  
        max_heap(vec, i, heap_size);  
    for(int i=heap_size; i>=1; i--)  
    {  
        SwapValue(vec[0],vec[i-1]);  
        max_heap(vec, 1, i);  
    }  
}  
void print(vector<int> vec)  
{  
    for(int i=0; i<vec.size(); i++)  
        cout<<vec[i]<<" ";  
    cout<<endl;  
}  
  
int main()  
{  
    vector<int> vec;  
    vec.push_back(23);  
    vec.push_back(5);  
    vec.push_back(1);  
    vec.push_back(10);  
    vec.push_back(13);  
    vec.push_back(32);  
    vec.push_back(21);  
    vec.push_back(14);  
    vec.push_back(19);  
    vec.push_back(20);  
      
    cout<<"排序前： "<<endl;  
    print(vec);  
      
    heapSort(vec);  
      
    cout<<"排序后： "<<endl;  
    print(vec);  
    return 0;  
}

2.求一个正整数N的开方，要求不能用库函数sqrt()，结果的精度在0.001
解析：牛顿迭代

#include<iostream>  
using namespace std;  
int main()  
{  
    int N;  
    cout<<"输入N的值：";  
    cin>>N  
  
    double x1 = 1;//初值  
    double x2 = x1/2.0+N/2.0/x1;  
    while( fabs(x2-x1)>0.001)  
    {  
        x1 = x2;  
        x2 = x1/2.0+N/2.0/x1;  
    }  
    cout<<x1<<endl;  
  
    return 0;  
}

3.给定一个矩阵intmaxtrixA[m][n],每行和每列都是增序的，实现一个算法去找矩阵中的某个元素element.

解法一：

#include<iostream>  
using namespace std;  
  
const int M = 4;  
const int N = 4;  
int main  
{  
    int matrix[M][N] = {};  
    double element;  
      
    int flag = 1;  
    for(int j=0; j<N; j++)  
    {  
        if(matrix[i][j] == element)  
            cout<<"位置"<<endl;  
        while( flag<M && matrix[i][j]<element )  
            --flag;  
        while( flag<M && matrix[i][j]>element )  
            ++flag;  
    }   
}

解法二：

bool Find(int *matrixA, int m, int n, int element)  
{  
    bool found = false;  
    if(matrixA != NULL & m & n)  
    {  
        int i,j;  
        i=0;j=n-1;  
        while(i<m;j>=0)  
        {  
            if(maxtrixA[i*n+j] == element)  
            {  
                found = true;  
                break;  
            }  
            else if(matrix[i*n+j]>element  
                --j;  
            else  
                ++i  
        }  
    }  
}

转载请注明原创链接：http://blog.csdn.net/wujunokay/article/details/12233289

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