Largest Rectangle in a Histogram(单调栈)

Largest Rectangle in a Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 33874		Accepted: 11045

Description

A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles: 

Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.

Input

The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that 1<=n<=100000. Then follow n integers h1,...,hn, where 0<=hi<=1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.

Output

For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.

Sample Input

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0

Sample Output

8
4000

   求：直方图中的最大矩形 

   思路：暴力法的话就是遍历所有矩形，求出每个矩形向左向右延伸的最大宽度，然后乘以高度求出最大值 O（n^2）

   单调栈：O（n）

  当新元素的高度大于栈顶元素，直接入栈

 否则，计算栈顶元素的最大扩展面积，取最大值，出栈

import java.util.Scanner;
import java.util.Stack;
class node{
	  long height;
	  int width;
}
public class Main {
	       static int n;
	       static Stack<node> st=new Stack<node>();
	       public static void main(String[] args) {
	           Scanner scan=new Scanner(System.in);
	           while(scan.hasNext()){
      	         n=scan.nextInt();
      	         if(n==0) break;
      	         long ans=0;
      	         for(int i=1;i<=n+1;i++){
      	        	     node a=new node();//每次new
      	    	         if(i<=n) a.height=scan.nextInt();
      	    	         else   a.height=0;
      	    	         a.width=i;
      	    	          while(!st.isEmpty() && st.peek().height>=a.height ){
      	    	        	     ans=Math.max(ans, st.peek().height*(i-st.peek().width));
      	    	        	     a.width=st.peek().width;
      	    	        	     st.pop();
      	    	          }
      	    	          st.push(a);
      	         }
      	        System.out.println(ans);
	           }
		}
}