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。

提示：双端 LIS 问题，用 DP 的思想可解。

#include <iostream>
using namespace std;

void printArray(int* arr, int len) {
	for (int i = 0; i < len; ++i) {
		cout << arr[i] << ' ';
	}
	cout << endl;
}

int binaryFind(int* lis, int target, int end) {
	int low = 0;
	int high = end;
	while (low <= high) {
		int mid = (low + high) / 2;
		if (target > lis[mid]) {
			low = mid + 1;
		} else if (target < lis[mid]) {
			high = mid - 1;
		} else {
			return mid;
		}
	}
	return low;
}

/*B[i]是从左到右的，0~i个数之间满足递增的数字个数；
 * C[i]为从右到左的，n- 1 ~ i个数之间满足递增的数字个数
 * 最后结果为n - max{B[i] + C[i] - 1}
 * */
int doubleEndLIS(int* arr, int len) {
	int* dp1 = new int[len];
	int* dp2 = new int[len];
	int* lis = new int[len];
	memset(dp1, 0, sizeof(int) * (len));
	memset(dp2, 0, sizeof(int) * (len));
	memset(lis, 0, sizeof(int) * (len));

	dp1[0] = 1;
	lis[0] = arr[0];
	int maxlen = 1;
	for (int i = 1; i < len; ++i) {
		int pos = binaryFind(lis, arr[i], maxlen - 1);
		lis[pos] = arr[i];
		dp1[i] = pos + 1;
		if (pos >= maxlen) {
			maxlen++;
		}
	}

	memset(lis, 0, sizeof(int) * len);
	lis[0] = arr[len - 1];
	dp2[len - 1] = 1;
	maxlen = 1;
	for (int i = len - 2; i >= 0; --i) {
		int pos = binaryFind(lis, arr[i], maxlen - 1);
		lis[pos] = arr[i];
		dp2[i] = pos + 1;
		if (pos >= maxlen) {
			maxlen++;
		}
	}

//	printArray(dp1, len);
//	printArray(dp2, len);
	int mid = 0;
	int result = 0;
	for (int i = 0; i < len; ++i) {
		result = max(result, dp1[i] + dp2[i]);
		if (result == dp1[i] + dp2[i])
			mid = i;
	}

	result--;

//还原子序列
	lis = new int[result];
	memset(lis, 0, result * sizeof(int));
	int curLen = dp1[mid];

	for (int i = mid; i >= 0; --i) {
		if (dp1[i] == curLen) {
			lis[curLen - 1] = arr[i];
			curLen--;
			if (curLen == 0) {
				break;
			}
		}
	}

//结果集下标
	int index = dp1[mid];
	curLen = dp2[mid] - 1;
	for (int i = mid + 1; i < len; ++i) {
		if (dp2[i] == curLen) {
			lis[index++] = arr[i];
			curLen--;
			if (curLen == 0) {
				break;
			}
		}
	}

	printArray(lis, result);

	result = len - result;
	return result;
}