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

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

遍歷陣列，每找到一個區域性峰值，然後向前遍歷所有的值，算出共同的矩形面積，每次對比保留最大值，程式碼如下：

// Pruning optimize
class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int res = 0;
        for (int i = 0; i < height.size(); ++i) {
            if (i + 1 < height.size() && height[i] <= height[i + 1]) {
                continue;
            }
            int minH = height[i];
            for (int j = i; j >= 0; --j) {
                minH = min(minH, height[j]);
                int area = minH * (i - j + 1);
                res = max(res, area);
            }
        }
        return res;
    }
};
 

 也可用棧來維護一個高度遞增序列，每次遇到較小的高度就開始計算矩形面積，程式碼如下

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int res = 0;
        stack<int> st;
        heights.push_back(0);
        for (int i = 0; i < heights.size(); ++i) {
            while (!st.empty() && heights[st.top()] >= heights[i]) {
                int cur = st.top(); st.pop();
                res = max(res, heights[cur] * (st.empty() ? i : (i - st.top() - 1)));
            }
            st.push(i);
        }
        return res;
    }
};