創新工場2014筆試演算法題彙總
阿新 • • 發佈:2018-10-31
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<<"排序後: "<<endl; print(vec); return 0; }
2.求一個正整數N的開方,要求不能用庫函式sqrt(),結果的精度在0.001
解析:牛頓迭代
#include<iostream> using namespace std; int main() { int N; cout<<"輸入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.給定一個矩陣intmaxtrixA[m][n],每行和每列都是增序的,實現一個演算法去找矩陣中的某個元素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
}
}
}
轉載請註明原創連結:http://blog.csdn.net/wujunokay/article/details/12233289