一个Codility上的练习遇到和预期结果不同的问题。

前几天有空的时候挑战了一个AMBITIOUS的题目 Prime and composite number条目下面的Flags。

题目如下：

A non-empty zero-indexed array A consisting of N integers is given. A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N ? 1 and A[P ? 1] < A[P] > A[P + 1].

For example, the following array A:

    A[0] = 1 

    A[1] = 5 

    A[2] = 3 

    A[3] = 4 

    A[4] = 3 

    A[5] = 4 

    A[6] = 1 

    A[7] = 2 

    A[8] = 3 

    A[9] = 4 

    A[10] = 6 

    A[11] = 2

has exactly four peaks: elements 1, 3, 5 and 10.

You are going on a trip to a range of mountains whose relative heights are represented by array A, as shown in a figure below. You have to choose how many flags you should take with you. The goal is to set the maximum number of flags on the peaks, according
to certain rules.



Flags can only be set on peaks. What's more, if you take K flags, then the distance between any two flags should be greater than or equal to K. The distance between indices P and Q is the absolute value |P ? Q|.

For example, given the mountain range represented by array A, above, with N = 12, if you take:

two flags, you can set them on peaks 1 and 5;

three flags, you can set them on peaks 1, 5 and 10;

four flags, you can set only three flags, on peaks 1, 5 and 10.

You can therefore set a maximum of three flags in this case.

Write a function:

int solution(vector<int> &A);

that, given a non-empty zero-indexed array A of N integers, returns the maximum number of flags that can be set on the peaks of the array.

For example, the following array A:

    A[0] = 1 

    A[1] = 5 

    A[2] = 3 

    A[3] = 4 

    A[4] = 3 

    A[5] = 4 

    A[6] = 1 

    A[7] = 2 

    A[8] = 3 

    A[9] = 4 

    A[10] = 6 

    A[11] = 2

the function should return 3, as explained above.

Assume that:

N is an integer within the range [1..200,000];

each element of array A is an integer within the range [0..1,000,000,000].

Complexity:

expected worst-case time complexity is O(N);

expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

如下是网上的其他人的实现结果：

// you can use includes, for example:
#include <algorithm>
#include <cmath>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
// write your code in C++11
int N = A.size();
vector<int> B(N,0);
int pos = 0;
int num = 0;
int res = 0;
for(int i=1;i<N-1;++i)
{
if((A[i]>A[i-1])&&(A[i]>A[i+1]))
B[i]=1;
}
vector<int> C(N,-1);
//C[N-1] = -1;//C[N-2] = -1;
for(int i=N-2;i>-1;--i)
{
if(B[i])
C[i] = i;
else
C[i] = C[i+1];
}
for(int i=1;i<=sqrt(N)+1;++i)
{
pos = 0;
num = 0;
while(pos<N&&num<i)
{
pos = C[pos];
if(pos == -1)
break;
pos += i;
num++;
}
res = max(res,num);
}
return res;
}

我自己开始的时候形成了思路实现方式和上面的方式非常像，但是总是不能得到100%的结果检验的正确率。
int solution(vector<int> &A) {
// write your code in C++11
vector<int> peaks;
int result=0;
for(unsigned int i=1;i<A.size()-1;i++)
{
if((A[i]>A[i-1]) && (A[i]>A[i+1]))
{
peaks.push_back(i);
i++;
}
}

for (unsigned int k=1; (k<=peaks.size())&&(k*k<A.size()); k++)
{
int flags=1;
int posA=0;
int posB=1;
while((flags<=k) && (posB<peaks.size()))
{
if( peaks[posB]-peaks[posA] >= k )
{
flags++;
posA=posB;
posB++;
}
else
{
posB++;
}
}
result=max(flags,result);
}
return result;
}

得到的结果不能100%正确，思考了很久也不能确定问题出在哪里。下图是结果



求高手指点。