我试图检查每个问题的3个解决方案经过多少时间,但是有时我会遇到运行时错误,看不到第3个解决方案的经过时间,但有时它可以工作。我认为solutions.h文件是正确的,但无论如何我都将其放在这里。
#include <iostream>
#include <cstdlib>
#include <ctime>
#include "solutions.h"
using namespace std;
int main()
{
cout << "Hello world!" << endl;
int* input1 = new int[10000];
int* input2 = new int[20000];
int* input3 = new int[40000];
int* input4 = new int[80000];
int* input5 = new int[100000];
for(int i = 0; i<100000; i++)
{
input1[i]= rand();
input2[i]= rand();
input3[i]= rand();
input4[i]= rand();
input5[i]= rand();
}
int* output1= new int[1000];
double duration;
clock_t startTime1 = clock();
solution1(input1,10000,1000,output1);
duration = 1000 * double( clock() - startTime1 ) / CLOCKS_PER_SEC;
cout << "Solution 1 with 10000 inputs took " << duration << " milliseconds." << endl;
startTime1 = clock();
solution2(input1,10000,1000,output1);
duration = 1000 * double( clock() - startTime1 ) / CLOCKS_PER_SEC;
cout << "Solution 2 with 10000 inputs took " << duration<< " milliseconds." << endl;
startTime1 = clock();
solution3(input1,10000,1000,output1);
duration = 1000 * double( clock() - startTime1 ) / CLOCKS_PER_SEC;
cout << "Solution 3 with 10000 inputs took " << duration << " milliseconds." << endl<<endl<<endl;
return 0;
}
并且solutions.h是
#ifndef SOLUTIONS_H_INCLUDED
#define SOLUTIONS_H_INCLUDED
#include <cmath>
void solution1( int input[], const int n, const int k, int output[] );
void solution2( int input[], const int n, const int k, int output[] );
void solution3( int input[], const int n, const int k, int output[] );
void swap( int &n1, int &n2 ) {
int temp = n1;
n1 = n2;
n2 = temp;
}
void solution1( int input[], const int n, const int k, int output[] ) {
int maxIndex, maxValue;
for( int i = 0; i < k; i++ ) {
maxIndex = i;
maxValue = input[i];
for( int j = i+1; j < n; j++ ) {
if( input[j] >= maxValue ) {
maxIndex = j;
maxValue = input[ j ];
}
}
swap( input[i], input[maxIndex] );
output[i] = input[i];
}
}
int partition( int input[], int p, int r ) {
int x = input[ r ], i = p - 1;
for( int j = p; j < r; j++ ) {
if( input[ j ] >= x ) {
i = i + 1;
swap( input[i], input[j] );
}
}
swap( input[i+1], input[r] );
return i + 1;
}
void quickSort( int input[], int p, int r ) {
int q;
if( p < r ) {
q = partition( input, p, r );
quickSort( input, p, q - 1 );
quickSort( input, q + 1, r );
}
}
void solution2( int input[], const int n, const int k, int output[] ) {
quickSort( input, 0, n - 1 );
for( int i = 0; i < k; i++ ) {
output[i] = input[i];
}
}
int partition2( int input[], int a, int p, int r ) {
int x = a, i = p - 1;
for( int j = p; j < r; j++ ) {
if( input[ j ] == x ) {
swap( input[ j ], input[ r ] );
}
if( input[ j ] >= x ) {
i = i + 1;
swap( input[i], input[j] );
}
}
swap( input[ i + 1 ], input[ r ] );
return i + 1;
}
void quickSort2( int input[], int p, int r ) {
int q;
if( p < r ) {
q = partition2( input, input[ r ], p, r );
quickSort2( input, p, q - 1 );
quickSort2( input, q + 1, r );
}
}
int findMin( int n1, int n2 ) {
if( n1 <= n2 )
return n1;
else
return n2;
}
int select( int input[], int n, int k, int start, int end, int flag ) {
if( n <= 5 ) {
quickSort2( input, start, end );
return input[ start + k - 1 ];
}
int i = start, numGroups = (int) ceil( ( double ) n / 5 ), numElements, j = 0;
int *medians = new int[numGroups];
while( i <= end ) {
numElements = findMin( 5, end - i + 1 );
medians[( i - start ) / 5] = select( input, numElements, (int) ceil( ( double ) numElements / 2 ), i, i + numElements - 1, 1 );
i = i + 5;
}
int M = select( medians, numGroups, (int) ceil( ( double ) numGroups / 2 ), 0, numGroups - 1, 1 );
delete[] medians;
if( flag == 1 )
return M;
int q = partition2( input, M, start, end );
int m = q - start + 1;
if( k == m )
return M;
else if( k < m )
return select( input, m - 1, k, start, q - 1, 0 );
else
return select( input, end - q, k - m, q + 1, end, 0 );
}
void solution3( int input[], const int n, const int k, int output[] ) {
select( input, n, k, 0, n - 1, 0 );
for( int i = 0; i < k; i++ )
output[i] = input[i];
}
#endif // SOLUTIONS_H_INCLUDED
使用地址清理器(clang ++ clock.cxx -std = c ++ 11 -O1 -g -fsanitize = address -fno-omit-frame-pointer)构建程序会发现以下问题:
$ ./a.out
Hello world!
=================================================================
==8175==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x62e00000a040 at pc 0x000104dbd912 bp 0x7fff5ae43970 sp 0x7fff5ae43968
WRITE of size 4 at 0x62e00000a040 thread T0
#0 0x104dbd911 in main clock.cxx:18
#1 0x7fff88cd85fc in start (libdyld.dylib+0x35fc)
#2 0x0 (<unknown module>)
0x62e00000a040 is located 0 bytes to the right of 40000-byte region [0x62e000000400,0x62e00000a040)
并且有您的代码:
int* input1 = new int[10000];
int* input2 = new int[20000];
int* input3 = new int[40000];
int* input4 = new int[80000];
int* input5 = new int[100000];
for(int i = 0; i<100000; i++)
{
input1[i]= rand();
input2[i]= rand();
input3[i]= rand();
input4[i]= rand();
input5[i]= rand();
}
如您所见,input1,input2,...,input4的大小是10K,20K,40K,80K元素,但是在循环中,我们正在访问该数组之外的元素,因此这可能导致堆损坏。
进程返回-1073741819(0xC0000005)
这意味着“内存访问冲突”或SEGFAULT。
希望这会有所帮助。
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