Largest Rectangle in Histogram, 求矩形图中最大的长方形面积

问题描述：

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.

算法分析：有两种方法，第一种暴力法，利用两重循环，求[i,j]之间最大的矩形面积。第二种方法利用栈，栈中存放递增的索引。

方法一：brute force

//brute force,用两重循环，求[i,j]之间最小值
	public static int largestRectangleArea(int[] heights)
	{
		int minHeight = 0;
		int maxArea = 0;
		for(int i = 0; i < heights.length; i ++)
		{
			for(int j = i; j < heights.length; j ++)
			{
				if(i == j)
				{
					minHeight = heights[i];
				}
				else
				{
					if(heights[j] < minHeight)
					{
						minHeight = heights[j];
					}
				}
				int temp = minHeight * (j - i + 1);
				if(maxArea < temp)
				{
					maxArea = temp;
				}
			}
		}
		return maxArea;
    }

 方法二：http://www.cnblogs.com/lichen782/p/leetcode_Largest_Rectangle_in_Histogram.html

//利用栈和单调性
	public static int largestRectangleArea2(int[] heights)
	{
		Stack<Integer> stack = new Stack<>();
		int[] h = Arrays.copyOf(heights, heights.length + 1);//h最后元素补0，为了让所有元素出栈，所以补0
		int i = 0;
		int maxArea = 0;
		while(i < h.length)
		{
			if(stack.isEmpty() || h[stack.peek()] <= h[i])
			{
				stack.push(i++);//只存放单调递增的索引
			}
			else
			{
				int t = stack.pop();//stack.isEmpty说明i是栈里最小的元素，面积为i*h[t]
				maxArea = Math.max(maxArea, h[t]*(stack.isEmpty() ? i : i-stack.peek()-1));
			}
		}
		return maxArea;
	}