問題描述：
比較給定直方圖對應的值列表爲[1, 2, 4, 2, 1]，那麼最大的矩形面積應該是[2, 4, 2]描述的區間矩形面積爲6(寬度爲3，高度爲2)。
關於這個問題，網上有比較多的解決方案，時間複雜度從O(n^2)、O(nlgn)，到O(n) (需O(n)空間)。
http://blog.csdn.net/arbuckle/article/details/710988
最蠻力的解決方案O(n^2)，就是查看所有的子直方圖，求解最大的矩形面積。
O(nlgn)的解決方案符合T(n) = 2 T(n/2) + O(n)的主方法。
O(n)的解決方案比較巧妙，可以從2種不同的角度來考慮，SO上有比較不錯的解答。其中比較容易理解的角度，可以參見這篇文章http://tech-queries.blogspot.com/2011/03/maximum-area-rectangle-in-histogram.html?1356186085，或者下面的解釋：
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For each bar, we must be able to find the biggest rectangle containing this bar. So the biggest one of these n rectangles is what we want.
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To get the biggest rectangle for a certain bar (let's say bar[i], the (i+1)th bar), we just need to find out the biggest interval that contains this bar. What we know is that all the bars in this interval must be at least the same height with bar[i]. So if we figure out how many consecutive same-height-or-higher bars are there on the immediate left of bar[i], and how many consecutive same-height-or-higher bars are there on the immediate right of the bar[i], we will know the length of the interval, which is the width of the biggest rectangle for bar[i].
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To count the number of consecutive same-height-or-higher bars on the immediate left of bar[i], we only need to find the closest bar on the left that is shorter than the bar[i], because all the bars between this bar and bar[i] will be consecutive same-height-or-higher bars.
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We use a stack to dynamicly keep track of all the left bars that are shorter than a certain bar. In other words, if we iterate from the first bar to bar[i], when we just arrive at the bar[i] and haven't updated the stack, the stack should store all the bars that are no higher than bar[i-1], including bar[i-1] itself. We compare bar[i]'s height with every bar in the stack until we find one that is shorter than bar[i], which is the cloest shorter bar. If the bar[i] is higher than all the bars in the stack, it means all bars on the left of bar[i] are higher than bar[i].
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We can do the same thing on the right side of the i-th bar. Then we know for bar[i] how many bars are there in the interval.
比較難懂的角度，解釋如下：
We process the elements in left-to-right order and maintain a stack of information about started but yet unfinished subhistograms. Whenever a new element arrives it is subjected to the following rules. If the stack is empty we open a new subproblem by pushing the element onto the stack. Otherwise we compare it to the element on top of the stack. If the new one is greater we again push it. If the new one is equal we skip it. In all these cases, we continue with the next new element. If the new one is less, we finish the topmost subproblem by updating the maximum area w.r.t. the element at the top of the stack. Then, we discard the element at the top, and repeat the procedure keeping the current new element. This way, all subproblems are finished until the stack becomes empty, or its top element is less than or equal to the new element, leading to the actions described above. If all elements have been processed, and the stack is not yet empty, we finish the remaining subproblems by updating the maximum area w.r.t. to the elements at the top.
For the update w.r.t. an element, we find the largest rectangle that includes that element. Observe that an update of the maximum area is carried out for all elements except for those skipped. If an element is skipped, however, it has the same largest rectangle as the element on top of the stack at that time that will be updated later. The height of the largest rectangle is, of course, the value of the element. At the time of the update, we know how far the largest rectangle extends to the right of the element, because then, for the first time, a new element with smaller height arrived. The information, how far the largest rectangle extends to the left of the element, is available if we store it on the stack, too.
We therefore revise the procedure described above. If a new element is pushed immediately, either because the stack is empty or it is greater than the top element of the stack, the largest rectangle containing it extends to the left no farther than the current element. If it is pushed after several elements have been popped off the stack, because it is less than these elements, the largest rectangle containing it extends to the left as far as that of the most recently popped element.
Every element is pushed and popped at most once and in every step of the procedure at least one element is pushed or popped. Since the amount of work for the decisions and the update is constant, the complexity of the algorithm is O(n) by amortized analysis.
def max_rectangle_area(histogram):
"""Find the area of the largest rectangle that fits entirely under
the histogram.
"""
stack = []
top = lambda: stack[-1]
max_area = 0
pos = 0 # current position in the histogram
for pos, height in enumerate(histogram):
start = pos # position where rectangle starts
while True:
if not stack or height > top().height:
stack.append(Info(start, height)) # push
elif stack and height < top().height:
max_area = max(max_area, top().height*(pos-top().start))
start, _ = stack.pop()
continue
break # height == top().height goes here
pos += 1
for start, height in stack:
max_area = max(max_area, height*(pos-start))
return max_area