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 = 0; // 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 = 0;
            while (i < n)
            {
                // If this bar is higher than the  
                // bar on top stack, push it to stack  
                if (s.Count == 0 || 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 == 0 ? i : i - s.Peek() - 1);

                    // 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 > 0)
            {
                tp = s.Peek();
                s.Pop();
                area_with_top = hist[tp] *
                               (s.Count == 0 ? i : i - s.Peek() - 1);

                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。