题目：

https://leetcode.com/problems/largest-rectangle-in-histogram/

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3]
.

The largest rectangle is shown in the shaded area, which has area = 10
unit.

For example,Given heights = [2,1,5,6,2,3],
return 10.

分析：

This question is to get a maximum area of largest rectangle in the histogram.

Input is an array.
Output is an int;

The brute force method is to scan its left side and right side for each bar, till the left boundary and right boundary which are lower than this bar.
Then the area would be its height times index of right boundary minus left boundary.
After all the bars are calculated, we can get the max area.

Another method is not to calculate the repeated work during the scanning.
For example, when we scan the right side of bar_index=2( height=5 ), its right boundary is bar_index=4( height=2 ).
And when we start to scan the left boundary for bar_index=4( height=2 ), in fact, we should already know bar_index=2( height=5 ) can cover this bar, so it will save time if we reduce the duplicate scanning, and store them in a data structure.

Here, we try to use stack to store such information.

1. We scan from left to right.
2. We add bar's index where might be the left boundary of its following bars, of course including current bar itself.
3. We stop adding index as soon as we meet a bar which is lower then the height of top in the stack, since it means we have found the right boundary of the newest bar in the stack.
4. Then we can start to calculate the possible area which are covered by the histogram determined by the stack.
5. The formula is easy: (right boundary index - left boundary index) * current height
   reminder:
   the right boundary index is where the stack is stopped.
   the nearest left boundary is just the one bar in front of current bar.
   the height of a bar can be calculated in the bar is which is higher than both of its left AND right boundary.

For detail explanation can be found in code.

Java

public class Solution {

    public int largestRectangleArea(int[] heights){
    
        if ( heights == null || heights.length == 0 )
            return 0;
        
        Deque<Integer> stack = new LinkedList<Integer>();   //store index
        int max = 0;
        
        for ( int i = 0; i <= heights.length; i++ ){
            
            int curVal = (i == heights.length) ? 0 : heights[i];                //如果i到array的外面了，curVal＝0，否则就为当前index装的高度
            
            while ( !stack.isEmpty() && heights[stack.peekLast()] >= curVal ){      //curVal 小于heights[stack.peekLast()]，说明cur是最顶端一点的右边界
                                                                                    //如果 curVal 大于stack.peekLast()，说明这是cur的一个左边界
                                                                                    //stack里每个元素都是后一个元素的左边界，停止add的时候说明碰到了右边界
                int height = heights[stack.pollLast()];                             //pollLast get height(index) AND REMOVE index from stack
                int leftBound = stack.isEmpty() ? 0 : (stack.peekLast() + 1);       //注意 why +1: since pollLast() removed
                int rightBound = i;
                max = Math.max( max, height * (rightBound - leftBound) );           //用公式计算面积
            }       

            stack.addLast(i);                           //stack=[1,4]时，说明 1处比4处值小，但是2处被弹出去了，说明4处曾经是2处的右边界，4处相当于凹心
            
        }
        
        
        return max;
    }
}