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在数组中删除尽可能少的数,使得数组满足“先由小到大,再由大到小”

2013-12-04 15:51 197 查看
void MyDelete(int* a, int n)
{
int* asc = new int
;    // asc[i]表示以a[i]结尾的最长升序数的个数
int* desc = new int
;   // desc[i]表示从a[i]开始的最长降序数的个数
int* inc = new int
;    // 亮点,用辅助数组的巧妙之处
const int max_int = numeric_limits<int>::max();

//求解asc[];
for(int i=0; i<n; ++i)
inc[i] = max_int;
for(int i=0; i<n; ++i)
{
int low = 0, high = i;
while(low <= high)
{
int mid = (low+high)/2;
if(inc[mid] <= a[i])
low = mid + 1;
else
high = mid - 1;
}
asc[i] = high+2;
inc[high+1] = a[i];
}
//求解desc[];
for(int i=0; i<n; ++i)
inc[i] = max_int;
for(int i=n-1; i>=0; --i)
{
int low = 0, high = n-1-i;
while(low <= high)
{
int mid = (low+high)/2;
if(inc[mid] <= a[i])
low = mid + 1;
else
high = mid - 1;
}
desc[i] = high + 2;
inc[high+1] = a[i];
}
// const int max_len = max(asc[i]+desc[i]-1);     0<= i < n
}

此题的巧妙之处在于,用辅助数据inc[]求解最长升序 数的个数。

题目变型: 求数组最长升序

int len = asc[0];
int index = 0;
for(int j=1; j<n; ++j)
if(len < asc[j])
{
len = asc[j];
index = j;
}
vector<int> vect;
for(int j=index; j>=0; --j)
{
if(len == asc[j])
{
vect.push_back(a[j]);
len--;
}
}
copy(vect.rbegin(),vect.rend(), ostream_iterator<int>(cout,"  "));
cout << endl;
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