[LeetCode]Largest Rectangle in Histogram

最新推荐文章于 2019-04-10 21:02:20 发布

原创最新推荐文章于 2019-04-10 21:02:20 发布 · 1.1k 阅读

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本文介绍了一种求解最大矩形面积的高效算法，通过使用栈来记录柱状图的高度和宽度，实现了从O(n^2)到O(n)的时间复杂度优化。文章提供了两种实现方式，包括详细的数据结构定义和关键步骤解析。

class Solution {
//first O(n^2)TLE: enumerate every bar i, look for the nearest left bar smaller than bar i, and nearest right bar smaller than bar i
//second O(n):using an stack to keep record of the height and width, the key observation: if h[j]<h[i](j>i), then position k(k>j) will never need 
//to touch h[i], only need to record the width of j
public:
	int largestRectangleArea(vector<int> &height) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int ans = 0;
		stack<pair<int,int>> s;//record height and width
		s.push(make_pair(0, 0));
		for (int i = 0; i < height.size()+1; ++i)
		{
			int curHeight;
			if(i < height.size())
				curHeight = height[i];
			else curHeight = -1;

			if(curHeight > s.top().first)//it is higher than the current top height
				s.push(make_pair(curHeight, 1));
			else
			{//if height is smaller than the current top height, than take down all higher stick in stack
			 //in the meantime we need to calculate the size 
				int minH = 0x7FFFFFFF;//assign the MAX_INT to min, test data contains the MAX_INT
				int sumWidth = 0;
				while (!s.empty() && s.top().first >= curHeight)
				{
					minH = min(minH, s.top().first);
					sumWidth += s.top().second;
					ans = max(minH*sumWidth, ans);
					s.pop();
				}
				s.push(make_pair(curHeight, sumWidth+1));
			}
		}
		return ans;
	}
};

second time

class Solution {
public:
    struct Node
    {
        int h;
        int l;
        Node(int _h = 0, int _l = 0):h(_h),l(_l){};
    };
    int largestRectangleArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        stack<Node> hs;
        int maxArea = 0;
        for(int i = 0; i < height.size(); ++i)
        {
            int curH = height[i];
            if(hs.empty()) hs.push(Node(curH, 1));
            else if(hs.top().h <= curH) hs.push(Node(curH, 1));
            else
            {
                int curL = 0;
                int minH = INT_MAX;
                while(!hs.empty() && hs.top().h > curH)
                {
                    curL += hs.top().l;
                    minH = hs.top().h;
                    maxArea = max(maxArea, curL*minH);
                    hs.pop();
                }
                hs.push(Node(curH, 1+curL));
            }
        }
        //
        int curL = 0;
        int minH;
        while(!hs.empty())
        {
            curL += hs.top().l;
            minH = hs.top().h;
            maxArea = max(maxArea, curL*minH);
            hs.pop();
        }
        return maxArea;
    }
};