从0~13中任取出7个数,然后判断这7个数中是否存在连续的5个数, 规则如下:
1) 7个数可以是重复的数.
2) 0可以表示任意数
例子如下:
0, 1, 4, 3, 8, 0, 13--->true: 1-2-3-4-5
0, 1, 1, 1, 9, 10, 0--->false
0, 1, 3, 9, 10, 11, 12->true: 9-10-11-12-13
0, 0, 0, 0, 0, 0, 0->true: 0-1-2-3-4
这是最近看到的一个算法题, 粗粗一看,觉得很简单, 可是慢慢往里面想,觉得要考虑的还是挺多的。现在把它实现出来放在这里,当然,加了几个参数使其更加通用。希望对大家有些参考价值。写的不明白的地方,有错误的地方大家可以指出来;大家如果有好的思路的话也希望能写下来共享一下。以下是代码与注释:
#include <stdio.h>
#include <iostream>
using namespace std;
/*
从0~13中任取出7个数,然后判断这7个数中是否存在连续的5个数, 规则如下:
1) 7个数可以是重复的数.
2) 0可以表示任意数
例子如下:
0, 1, 4, 3, 8, 0, 13--->true: 1-2-3-4-5
0, 1, 1, 1, 9, 10, 0--->false
0, 1, 3, 9, 10, 11, 12->true: 9-10-11-12-13
0, 0, 0, 0, 0, 0, 0->true: 0-1-2-3-4
*/
// Helper functions
void outputarray(int a[], int n)
{
for(int i = 0; i < n; ++i) cout<<a[i]<<" ";
cout<<endl;
}
void outputresult(int nstart, int m)
{
cout<<"Continuous Array:";
while(m--) cout<<nstart++<<" ";
cout<<endl;
}
// Return if the n elements array contains m continuous numbers, the elements must large than 0
// this is a common implementation
bool Is_n_Contains_m_ContinuousNum(int a[], int n, int m)
{
// step 1: get num of 0
int nZeroNum = 0;
for(int i = 0; i < n; ++i)
if(0 == a[i]) ++nZeroNum;
cout<<"Original Array: "; outputarray(a, n);
// step 2: if we have enough 0, get continuous num is easy.
if(nZeroNum >= m-1)
{
int min = a[0];
for(int i = 1; i < n; ++i)
if(a[i] < min || 0 == min) min = a[i];
outputresult(min, m);
return true;
}
// not enough zero, we need to refine the judgement
else
{
// step 2.1: sort the array. (bubble sort, ascending)
bool bIsDone = false;
for(int i = n-1; i >= 0 && !bIsDone; --i)
{
bIsDone = true;
for(int j = 0; j < i; ++j)
{
if(a[j+1] < a[j])
{
bIsDone = false;
int tmp = a[j+1];
a[j+1] = a[j];
a[j] = tmp;
}
}
}
cout<<"Sorted Array: "; outputarray(a, n);
// step 2.2: remove redundant num except 0
int aa[256];
aa[0] = a[0];
int j = 1;
for(int i = 0; i < n-1; ++i)
{
if(a[i+1] != a[i] || 0 == a[i+1])
aa[j++] = a[i+1];
}
memcpy(a, aa, j * sizeof(aa[0]));
n = j;
if(n < m) return false;
cout<<"Unique Array: "; outputarray(a, n);
// step 2.3: get index of first non-zero element
int nIndex = 0;
for(int i = 0; i < n; ++i)
{
if(a[i] != 0)
{
nIndex = i;
break;
}
}
// step 2.4: refined judgement
// if n = 7; m = 5; nZeroNum = 2;
// if we can get continious number without zero or only with 1 zero, then with 2 zero is ok too.
// so if we got x zeros, we need to check if can success with (0 ~ x-1) zeros
for(int k = 0; k <= nZeroNum; ++k)
{
int nInterval = m - k - 1;
for(int i = nIndex; i < n-nInterval; ++i)
{
// when k = nZeroNum = 2;
// if the a[i+nInterval] - a[i] ranged in (nInterval, m-1), then it is continuous)
// means a[i+2] - a[i] ranged in (2, 4), for example:
// 1 3 5; 1 2 3; 1 2 4;
if(a[i+nInterval] - a[i] <= m-1 && a[i+nInterval] - a[i] >= nInterval)
{
outputresult(a[i], m);
return true;
}
}
}
}
return false;
}
int main(int argc, char *argv[])
{
int a[] = {0, 0, 0, 0, 0, 0, 0};
if(!Is_n_Contains_m_ContinuousNum(a, sizeof(a)/sizeof(a[0]),5))
cout<<"Continuous Array:Not Found ";
return 0;
}
#include <iostream>
using namespace std;
/*
从0~13中任取出7个数,然后判断这7个数中是否存在连续的5个数, 规则如下:
1) 7个数可以是重复的数.
2) 0可以表示任意数
例子如下:
0, 1, 4, 3, 8, 0, 13--->true: 1-2-3-4-5
0, 1, 1, 1, 9, 10, 0--->false
0, 1, 3, 9, 10, 11, 12->true: 9-10-11-12-13
0, 0, 0, 0, 0, 0, 0->true: 0-1-2-3-4
*/
// Helper functions
void outputarray(int a[], int n)
{
for(int i = 0; i < n; ++i) cout<<a[i]<<" ";
cout<<endl;
}
void outputresult(int nstart, int m)
{
cout<<"Continuous Array:";
while(m--) cout<<nstart++<<" ";
cout<<endl;
}
// Return if the n elements array contains m continuous numbers, the elements must large than 0
// this is a common implementation
bool Is_n_Contains_m_ContinuousNum(int a[], int n, int m)
{
// step 1: get num of 0
int nZeroNum = 0;
for(int i = 0; i < n; ++i)
if(0 == a[i]) ++nZeroNum;
cout<<"Original Array: "; outputarray(a, n);
// step 2: if we have enough 0, get continuous num is easy.
if(nZeroNum >= m-1)
{
int min = a[0];
for(int i = 1; i < n; ++i)
if(a[i] < min || 0 == min) min = a[i];
outputresult(min, m);
return true;
}
// not enough zero, we need to refine the judgement
else
{
// step 2.1: sort the array. (bubble sort, ascending)
bool bIsDone = false;
for(int i = n-1; i >= 0 && !bIsDone; --i)
{
bIsDone = true;
for(int j = 0; j < i; ++j)
{
if(a[j+1] < a[j])
{
bIsDone = false;
int tmp = a[j+1];
a[j+1] = a[j];
a[j] = tmp;
}
}
}
cout<<"Sorted Array: "; outputarray(a, n);
// step 2.2: remove redundant num except 0
int aa[256];
aa[0] = a[0];
int j = 1;
for(int i = 0; i < n-1; ++i)
{
if(a[i+1] != a[i] || 0 == a[i+1])
aa[j++] = a[i+1];
}
memcpy(a, aa, j * sizeof(aa[0]));
n = j;
if(n < m) return false;
cout<<"Unique Array: "; outputarray(a, n);
// step 2.3: get index of first non-zero element
int nIndex = 0;
for(int i = 0; i < n; ++i)
{
if(a[i] != 0)
{
nIndex = i;
break;
}
}
// step 2.4: refined judgement
// if n = 7; m = 5; nZeroNum = 2;
// if we can get continious number without zero or only with 1 zero, then with 2 zero is ok too.
// so if we got x zeros, we need to check if can success with (0 ~ x-1) zeros
for(int k = 0; k <= nZeroNum; ++k)
{
int nInterval = m - k - 1;
for(int i = nIndex; i < n-nInterval; ++i)
{
// when k = nZeroNum = 2;
// if the a[i+nInterval] - a[i] ranged in (nInterval, m-1), then it is continuous)
// means a[i+2] - a[i] ranged in (2, 4), for example:
// 1 3 5; 1 2 3; 1 2 4;
if(a[i+nInterval] - a[i] <= m-1 && a[i+nInterval] - a[i] >= nInterval)
{
outputresult(a[i], m);
return true;
}
}
}
}
return false;
}
int main(int argc, char *argv[])
{
int a[] = {0, 0, 0, 0, 0, 0, 0};
if(!Is_n_Contains_m_ContinuousNum(a, sizeof(a)/sizeof(a[0]),5))
cout<<"Continuous Array:Not Found ";
return 0;
}
PS:网友建议
fflush:不需要这么复杂吧,题目说明了是从0~13中任取出7个数,那么建立一个int A[13],记录哪些数有哪些数没有(有的话置A[i]为1,否则是0),然后检测A中连续的5个位置的情况就可以了
SmartPtr:fflush的思路对我很有启发, 但我们还要考虑多个0的情况, 按着这个思路的话我觉得可以这么做:针对0~13建立一个数组A[14], A[0], A[1]...分别对应0, 1...的个数。然后依次检测A中连续的5个位置, 如果其0的个数小于A[0],那么就存在连续的数。(当然还有一些边缘情况要处理)。
这个算法我觉得十分有效, 通过引入一个数组大大的简化了问题。但是有个缺点就是如果要判断任意范围内的5个连续数就不容易了。如:
0, 1, 122, 678, 10000, 3, 6
(搬自以前blog, 2007-08-26)