算法题（直方图中最大的矩形区域|S2） _最大矩形区域

本文概述

C ++
Java
Python3
C#

在给定的直方图中找到最大的矩形区域, 其中最大的矩形可以由许多连续的条形组成。为简单起见, 假定所有条形都具有相同的宽度, 并且宽度为1个单位。
例如, 考虑以下具有7个高度{6, 2, 5, 4, 4, 5, 1, 6}的条形图。可能的最大矩形是12(请参见下图, 最大面积的矩形以红色突出显示)

文章图片
我们已经讨论了基于分治法的O(nLogn)解决方案
对于这个问题。在这篇文章中, 讨论了O(n)时间解。像以前的帖子, 为简单起见, 所有条的宽度假定为1。对于每个" x"条, 我们以" x"作为矩形中最小的条计算面积。如果我们为每个小节" x"计算这样的面积并找到所有面积的最大值, 那么我们的任务就完成了。如何以" x"作为最小条形来计算面积？我们需要知道" x"左侧第一个较小(小于" x")条的索引, 以及" x"右侧第一个较小的索引。让我们将这些索引分别称为"左索引"和"右索引"。
我们从左到右遍历所有条, 并保持一堆条。每个条被推入堆栈一次。当看到较小高度的条时, 将从堆栈中弹出条。弹出条时, 我们以弹出条为最小条来计算面积。我们如何获得弹出栏的左右索引-当前索引告诉我们"右索引", 而堆栈中上一项的索引就是"左索引"。以下是完整的算法。
1)创建一个空堆栈。
2)从第一个小节开始, 然后对每个小节" hist [i]"执行跟踪, 其中" i"的范围从0到n-1。
……a)如果堆栈为空或hist [i]高于堆栈顶部的条, 则按" i"进行堆栈。
……b)如果此栏小于堆栈顶部, 则在堆栈顶部较大的同时, 继续移除堆栈顶部。令删除的条为hist [tp]。用hist [tp]作为最小条形来计算矩形的面积。对于hist [tp], "左索引"是堆栈中的上一个项目(在tp之前), "右索引"是" i"(当前索引)。
3)如果堆栈不是空的, 则一个接一个地从堆栈中删除所有条, 并对每个已删除的条执行步骤2.b。
以下是上述算法的实现。
C ++

//C++ program to find maximum rectangular area in //linear time #include< bits/stdc++.h> using namespace std; //The main function to find the maximum rectangular //area under given histogram with n bars int getMaxArea( int hist[], int n) { //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; 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.empty() || hist[s.top()] < = 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.top(); //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.empty() ? i : i - s.top() - 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.empty() == false ) { tp = s.top(); s.pop(); area_with_top = hist[tp] * (s.empty() ? i : i - s.top() - 1); if (max_area < area_with_top) max_area = area_with_top; }return max_area; }//Driver program to test above function int main() { int hist[] = {6, 2, 5, 4, 5, 1, 6}; int n = sizeof (hist)/sizeof (hist[0]); cout < < "Maximum area is " < < getMaxArea(hist, n); return 0; }

Java

//Java program to find maximum rectangular area in linear timeimport java.util.Stack; public class RectArea { //The main function to find the maximum rectangular area under given //histogram with n bars static int getMaxArea( int hist[], int n) { //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< Integer> s = new Stack< > (); 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.empty() || 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.empty() ? 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.empty() == false ) { tp = s.peek(); s.pop(); area_with_top = hist[tp] * (s.empty() ? i : i - s.peek() - 1 ); if (max_area < area_with_top) max_area = area_with_top; }return max_area; }//Driver program to test above function public static void main(String[] args) { int hist[] = { 6 , 2 , 5 , 4 , 5 , 1 , 6 }; System.out.println( "Maximum area is " + getMaxArea(hist, hist.length)); } } //This code is Contributed by Sumit Ghosh

Python3

# Python3 program to find maximum # rectangular area in linear timedef max_area_histogram(histogram):# This function calulates maximum # rectangular area under given # histogram with n bars# Create an empty stack. The stack # holds indexes of histogram[] list. # The bars stored in the stack are # always in increasing order of # their heights. stack = list ()max_area = 0 # Initialize max area# Run through all bars of # given histogram index = 0 while index < len (histogram):# If this bar is higher # than the bar on top # stack, push it to stackif ( not stack) or (histogram[stack[ - 1 ]] < = histogram[index]): stack.append(index) index + = 1# 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 : # pop the top top_of_stack = stack.pop()# Calculate the area with # histogram[top_of_stack] stack # as smallest bar area = (histogram[top_of_stack] * ((index - stack[ - 1 ] - 1 ) if stack else index))# update max area, if needed max_area = max (max_area, area)# Now pop the remaining bars from # stack and calculate area with # every popped bar as the smallest bar while stack:# pop the top top_of_stack = stack.pop()# Calculate the area with # histogram[top_of_stack] # stack as smallest bar area = (histogram[top_of_stack] * ((index - stack[ - 1 ] - 1 ) if stack else index))# update max area, if needed max_area = max (max_area, area)# Return maximum area under # the given histogram return max_area# Driver Code hist = [ 6 , 2 , 5 , 4 , 5 , 1 , 6 ] print ( "Maximum area is" , max_area_histogram(hist))# This code is contributed # by Jinay Shah

//C# program to find maximum //rectangular area in linear time using System; using System.Collections.Generic; class GFG { //The main function to find the //maximum rectangular area under //given histogram with n bars public static int getMaxArea( int [] hist, int n) { //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; }//Driver Code public static void Main( string [] args) { int [] hist = new int [] {6, 2, 5, 4, 5, 1, 6}; Console.WriteLine( "Maximum area is " + getMaxArea(hist, hist.Length)); } }//This code is contributed by Shrikant13

输出如下：

Maximum area is 12

时间复杂度：由于每个小节仅被推动和弹出一次, 因此此方法的时间复杂度为O(n)。
参考文献
http://www.informatik.uni-ulm.de/acm/Locals/2003/html/histogram.html
http://www.informatik.uni-ulm.de/acm/Locals/2003/html/judge.html
【算法题（直方图中最大的矩形区域|S2）】谢谢艾希什·阿南德(Ashish Anand)用于建议初始解决方案。如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。