01-复杂度2 Maximum Subsequence Sum

 思路

比上一题（01-复杂度1 最大子列和问题），要多记录几个内容

首尾元素，当前子序列开头元素，当前子序列结尾元素，最佳子序列开头元素，current是否在此处清零（则下一个元素是开头元素）

时间复杂度	O(n)
空间复杂度	O(1)

code(C++)

# include 
# include 


int main(void)
{
	long long ans = -1;
	long long current = 0;

	long long start, end, current_start;
	bool expectNext = true;

	int K;
	std::cin >> K;
	int firstNum, lastNum;
	int k = K;

	while (k--)
	{

		long long tmp;
		std::cin >> tmp;

		if (k == 0) lastNum = tmp;
		if (k == K - 1) firstNum = tmp;

		if (expectNext)
		{
			current_start = tmp;
			expectNext = false;
		}

		current += tmp;

		if (current < 0)
		{
			current = 0;
			expectNext = true;
		}
		else if (current > ans)
		{
			ans = current;
			start = current_start;
			end = tmp;
		}
	}
	if (ans >= 0)
		std::cout << ans << " " << start <<  " " << end << std::endl;
	else
		std::cout << 0 << " " << firstNum << " " << lastNum << std::endl;
	return 0;
}