[LeetCode OJ] Largest Rectangle in Histogram

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given height = [2,1,5,6,2,3],
return 10.

方法一：两层循环遍历，复杂度O（n²）

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int maxArea=0;
        for(unsigned i=0; i<height.size(); i++)
        {
            int min = height[i];
            for(unsigned j=i; j<height.size(); j++)
            {
                if(height[j]<min)
                    min = height[j];
                int area = min*(j-i+1);
                if(area>maxArea)
                    maxArea = area;
            }
        }
        return maxArea;
    }
};

方法二：用堆栈保存重要位置，复杂度O（n）

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {  //用堆栈来实现
        stack<unsigned> st;
        unsigned maxArea = 0;
        for(unsigned i=0; i<height.size(); i++)
        {
            if(st.empty())
                st.push(i);
            else
            {
                while(!st.empty())
                {
                    if(height[i]>=height[st.top()])
                    {
                        st.push(i);
                        break;
                    }
                    else
                    {
                        unsigned idx=st.top();
                        st.pop();
                        unsigned leftwidth = st.empty() ? idx : (idx-st.top()-1);
                        unsigned rightwidth = i - idx-1;
                        maxArea = max(maxArea, height[idx]*(leftwidth+rightwidth+1));
                    }
                }
                if(st.empty())
                    st.push(i);
            }
        }
        unsigned rightidx = height.size();
        while(!st.empty())
        {
            unsigned idx = st.top();
            st.pop();
            unsigned leftwidth = st.empty() ? idx : (idx-st.top()-1);
            unsigned rightwidth = rightidx - idx-1;
            maxArea = max(maxArea, height[idx]*(leftwidth+rightwidth+1));
        }
        return maxArea;
    }
};