leetcode84

 public class Solution
     {
         public int LargestRectangleArea(int[] hist)
         {
             // The main function to find the
             // maximum rectangular area under
             // given histogram with n bars
             int n = hist.Length;
             // Create an empty stack. The stack
             // holds indexes of hist[] array
             // The bars stored in stack are always
             // in increasing order of their heights.
             Stack<int> s = new Stack<int>();

             int max_area = ; // Initialize max area
             int tp; // To store top of stack
             int area_with_top; // To store area with top
                                // bar as the smallest bar 

             // Run through all bars of
             // given histogram
             int i = ;
             while (i < n)
             {
                 // If this bar is higher than the
                 // bar on top stack, push it to stack
                 if (s.Count ==  || hist[s.Peek()] <= hist[i])
                 {
                     s.Push(i++);
                 }

                 // If this bar is lower than top of stack,
                 // then calculate area of rectangle with
                 // stack top as the smallest (or minimum
                 // height) bar. 'i' is 'right index' for
                 // the top and element before top in stack
                 // is 'left index'
                 else
                 {
                     tp = s.Peek(); // store the top index
                     s.Pop(); // pop the top 

                     // Calculate the area with hist[tp]
                     // stack as smallest bar
                     area_with_top = hist[tp] *
                                    (s.Count ==  ? i : i - s.Peek() - );

                     // update max area, if needed
                     if (max_area < area_with_top)
                     {
                         max_area = area_with_top;
                     }
                 }
             }

             // Now pop the remaining bars from
             // stack and calculate area with every
             // popped bar as the smallest bar
             while (s.Count > )
             {
                 tp = s.Peek();
                 s.Pop();
                 area_with_top = hist[tp] *
                                (s.Count ==  ? i : i - s.Peek() - );

                 if (max_area < area_with_top)
                 {
                     max_area = area_with_top;
                 }
             }

             return max_area;
         }
     }

参考：https://www.geeksforgeeks.org/largest-rectangle-under-histogram/

补充一个python的实现：

 class Solution:
     def largestRectangleArea(self, heights: 'List[int]') -> int:
         heights.append(0)#默认最后补充一个0，方便统一处理
         n = len(heights)
         s = []
         max_area = 0#最大面积
         tp = 0#栈顶索引
         area_with_top = 0#临时面积
         i = 0
         while i < n:
             if len(s) == 0 or heights[s[-1]] <= heights[i]:
                 s.append(i)#栈内记录的是高度递增的索引
                 i += 1
             else:#遇到了高度比当前栈顶元素低的元素时，
                 tp = s.pop(-1)#清算栈内的元素的高度
                 if len(s) == 0:
                     area_with_top = heights[tp] * i#栈内没有元素，则宽度是i
                 else:#高度是栈顶元素，宽度是i - 1 - 前一个栈顶元素的索引
                     area_with_top = heights[tp] * (i - s[-1] - 1)
                 max_area = max(max_area,area_with_top)#更新最大值
         # while len(s) > 0:#处理栈内剩余元素，处理流程和遇到一个
         #     tp = s.pop(-1)
         #     if len(s) == 0:
         #         area_with_top = heights[tp] * i
         #     else:
         #         area_with_top = heights[tp] * (i - s[-1] - 1)
         #     max_area = max(max_area,area_with_top)
         return max_area

采用了一个小技巧，在heights最后补一个0，则21行到27行的代码就可以省略了。