从一列数中筛除尽可能少的数使得从左往右看，这些数是从小到大再从大到小的（网易）。

题目描述：

　　从一列数中筛除尽可能少的数使得从左往右看，这些数是从小到大再从大到小的（网易）。

分析：

　　这可以用双端LIS方法来解决，先求一遍从左到右的，再求一遍从右到左的。最后从里面选出和最大的即可。

代码实现：

#include <iostream>

using namespace std;

int DoubleEndLIS(int *arr, int len)
{
int *LIS = new int[len];
int *lefToRight = new int[len];        //leftToRight[i]表示0~i最长子序列长度-1
int *rightToLeft = new int[len];
int maxLen = 0;        //记录总共的(上升+下降)最长子序列长度
int low, high, mid;

for (int i = 0; i < len; ++i)
{
lefToRight[i]  = 0;
LIS[i] = 0;
}

LIS[0] = arr[0];
for (int i = 1; i < len; i++)
{
low = 0; high = lefToRight[i-1];
while (low <= high)
{
mid = (low + high)/2;
if (LIS[mid] < arr[i])
{
low = mid + 1;
}
else
{
high = mid -1;
}
}

LIS[low] = arr[i];
if (low > lefToRight[i-1])
{
lefToRight[i] = lefToRight[i-1] + 1;    //最长子序列长度加1
}
else
{
lefToRight[i] = lefToRight[i-1];
}
}

//leftToRight的每个值增加1，因为他们是最长子序列值-1
//此时leftToRight表示的是最长子序列的真正值。
for (int i = 0; i < len; i++)
{
lefToRight[i]++;
}

//从右到左
for (int i = 0; i < len; i++)
{
rightToLeft[i] = 0;
LIS[i] = 0;
}

int k = 0;
LIS[0] = arr[len-1];
for (int i = len -2; i >= 0; --i)
{
low = 0; high = rightToLeft[k];
while (low <= high)
{
mid = (low + high)/2;
if (LIS[mid] < arr[i])
{
low = mid + 1;
}
else
{
high = mid - 1;
}
}

LIS[low] = arr[i];
if (low > rightToLeft[k])
{
rightToLeft[k+1] = rightToLeft[k] + 1;
}
else
{
rightToLeft[k+1] = rightToLeft[k];
}
++k;
}

for (int i = 0; i < k; ++i)
{
rightToLeft[i]++;
}

//求最大值即为要求的
for (int i = 0; i < len; ++i)
{
cout<<"i: "<<i<<" "<<lefToRight[i]<<"  "<<rightToLeft[len-i-1]<<endl;
if (lefToRight[i] + rightToLeft[len-i-1] > maxLen)
maxLen = lefToRight[i] + rightToLeft[len-i-1];
}
cout<<"maxLen:"<<maxLen<<endl;

delete LIS;
delete lefToRight;
delete rightToLeft;

return len - maxLen + 1;
}

int main()
{
int arr[] = {1,5,7,6,9,3,8,4,2};
int ret;
ret = DoubleEndLIS(arr, 9);
cout<<ret<<endl;

return 0;
}

参考：http://blog.csdn.net/nciaebupt/article/details/8466049

但是，他的程序有问题，我做了修改。