Problem:
题目

Solve:
主要思想：用一个栈来保存高度的索引，当当前索引的高度比前一个大的时候就加入栈中，当比它小时，就从栈中不断的弹出索引，计算最大矩形的值，直到当前索引的高度比栈顶索引高度大时停止，并把当前索引加入到栈中。

[Java Code]

public int largestRectangleArea(int[] height) {
     Stack<Integer> stack = new Stack<Integer>();
        int n = height.length, area = 0;
        for (int i = 0; i < n; i++) {
            while (!stack.empty() && height[stack.peek()] > height[i]) {
                int h = height[stack.peek()];
                stack.pop();
                int l = stack.empty() ? -1 : stack.peek();
                area = Math.max(area, h * (i - l - 1));
            }
            stack.push(i);
        }
        while (!stack.empty() && height[stack.peek()] > 0) {
            int h = height[stack.peek()];
            stack.pop();
            int l = stack.empty() ? -1 : stack.peek();
            area = Math.max(area, h * (height.length - 1 - l));
        }
        return area;   
}