美丽塔O(n)解法单调栈

题目

本文涉及的基础知识点

单调栈分类、封装和总结https://blog.csdn.net/he_zhidan/article/details/135150163

时间复杂度

O(n)

分析

接着上篇。从左向右依次处理Left，处理Left[i]时，从右向左寻找第一个符合maxHeights[j]=maxHeights[j2]，那j1永远不会被选到。比如：{1,3,2,4,5}，由于2在3右边，且小于3，则无论如何不会选中3。{1,2,2.....}，后面无论有什么数，都不会选中第一个2，要么是其他数，要么是第二个2。
可以用栈实现，入栈maxHeights[i]之前，先出栈大于等于maxHeights[i]的数，剩余的都小于maxHeights[i]的数。也就是栈按升序排序的。由于maxHeights[i]和heights[i]都可以通过索引查询，栈中只需要记录索引。
Right类似，不再累赘。

样例分析

maxHeights	Left的栈情况
{1,2,3,4,5}	1 12 123 1234 12345
{5,4,3,2,1}	5 4 3 2 1
{1,2,4,3,5}	1 12 124 123 1235
{3,1,2}	3 1 12
{2,1,3}	2 1 13

代码

核心代码


class Solution {
public:
    long long maximumSumOfHeights(vector<int>& maxHeights) {
        m_c = maxHeights.size();
        m_vLeft.resize(m_c);
        m_vRight.resize(m_c);
        {//处理左边
            stack<int> sta;//记录做边的索引
            for (int i = 0; i < m_c; i++)
            {
                const auto& h = maxHeights[i];
                while (sta.size() && (maxHeights[sta.top()] >= h))
                {
                    sta.pop();//左边比右边大，不会被选中
                }
                if (sta.size())
                {
                    m_vLeft[i] = m_vLeft[sta.top()] + (long long)h * (i - sta.top());
                }
                else
                {
                    m_vLeft[i] =  (long long)h * (i -(-1) );
                }
                sta.emplace(i);
            }
        }
 
        {//处理右边
            stack<int> sta;//记录做边的索引
            for (int i = m_c - 1; i >= 0; i--)
            {
                const auto& h = maxHeights[i];
                while (sta.size() && (maxHeights[sta.top()] >= h))
                {
                    sta.pop();//左边比右边大，不会被选中
                }
                if (sta.size())
                {
                    m_vRight[i] = m_vRight[sta.top()] + (long long)h * (sta.top()-i);
                }
                else
                {
                    m_vRight[i] = (long long)h * (m_c-i);
                }
                sta.emplace(i);
            }
        }
        
        long long llRet = 0;
        for (int i = 0; i < m_c; i++)
        {//假定i是山顶            
            long long llCur = m_vLeft[i] + m_vRight[i] - maxHeights[i];
            llRet = max(llRet, llCur);
        }
        return llRet;
    }
    int m_c;
    vector<long long> m_vLeft, m_vRight;
};

测试用代码

class CDebug : public Solution
{
public:
   long long maximumSumOfHeights(vector& maxHeights, vector& vLeft, vector& vRight)
   {
       vector maxs(maxHeights.begin(), maxHeights.end());
       long long llRet = Solution::maximumSumOfHeights(maxs);
       for (int i = 0 ; i < vLeft.size();i++ )
       {
           assert(m_vLeft[i] == vLeft[i]);
           assert(m_vRight[i] == vRight[i]);
       }

       //调试用代码
       std::cout << "Left: ";
       for (int i = 0; i < m_c; i++)
       {
           std::cout << m_vLeft[i] << " ";
       }
       std::cout << std::endl;
       std::cout << "Right: ";
       for (int i = 0; i < m_c; i++)
       {
           std::cout << m_vRight[i] << " ";
       }
       std::cout << std::endl;
       return llRet;
   }
};
int main()
{
   vector < vector>> param = { {{1,2,3,4,5} ,{1,3,6,10,15},{5,8,9,8,5}} ,
       {{5,4,3,2,1},{5,8,9,8,5},{15,10,6,3,1}} ,
       {{1,2,4,3,5},{1,3,7,9,14},{5,8,10,6,5}},
   {{3,1,2}, {3,2,4},{5,2,2}},
   {{2,1,3},{2,2,5},{4,2,3}},
       {{1000000000,1000000000,1000000000},{1000000000,2000000000,3000000000LL},{3000000000LL,2000000000,1000000000}} };
   for (auto& vv : param)
   {
       auto res = CDebug().maximumSumOfHeights(vv[0], vv[1], vv[2]);
   }
   //auto res = Solution().maxPalindromes("rire", 3);

//CConsole::Out(res);
}

使用封装类的代码


class CRangIndex
{
public:
	template<class _Pr>
	CRangIndex(int iVectorSize, _Pr CurIndexCmpStackTopIndex)
	{
		m_c = iVectorSize;
		m_vLeft.assign(m_c, -1);
		m_vRight.assign(m_c, m_c);
		stack<int> sta;
		for (int i = 0; i < m_c; i++)
		{
			while (sta.size() && (CurIndexCmpStackTopIndex(i, sta.top())))
			{
				m_vRight[sta.top()] = i;
				sta.pop();
			}
			if (sta.size())
			{
				m_vLeft[i] = sta.top();
			}
			sta.emplace(i);
		}
	}
 
	template<class _Pr>
	CRangIndex(const vector<int>& nums, _Pr CurValueCmpStackTopValue)
	{
		m_c = nums.size();
		m_vLeft.assign(m_c, -1);
		m_vRight.assign(m_c, m_c);
		stack<int> sta;
		for (int i = 0; i < m_c; i++)
		{
			while (sta.size() && (CurValueCmpStackTopValue(nums[i], nums[sta.top()])))
			{
				m_vRight[sta.top()] = i;
				sta.pop();
			}
			if (sta.size())
			{
				m_vLeft[i] = sta.top();
			}
			sta.emplace(i);
		}
	}
	int m_c;
	vector<int> m_vLeft, m_vRight;//vLeft[i] 从右向左第一个小于nums[i] ;vRight[i] 是第一个小于等于nums[i]。
};
 
class Solution {
public:
	long long maximumSumOfHeights(vector<int>& maxHeights) {
		CRangIndex ri(maxHeights, std::less<>());
		m_vLeft.resize(ri.m_c);
		for (int i = 0; i < ri.m_c; i++)
		{
			m_vLeft[i] = (-1 == ri.m_vLeft[i]) ? 0 : m_vLeft[ri.m_vLeft[i]];
			m_vLeft[i] += ((long long)i - ri.m_vLeft[i]) * maxHeights[i];
		}		
		m_vRight.resize(ri.m_c);
		long long llRet = 0;
		for (int i = ri.m_c - 1; i >= 0; i--)
		{
			m_vRight[i] = (ri.m_c == ri.m_vRight[i]) ? 0 : m_vRight[ri.m_vRight[i]];
			m_vRight[i] += ((long long)ri.m_vRight[i] - i ) * maxHeights[i];
			long long llCur = m_vLeft[i] + m_vRight[i] - maxHeights[i];
			llRet = max(llRet, llCur);
		}
		return llRet;
	}
	int m_c;
	vector<long long> m_vLeft, m_vRight;
};