84. 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.

思路：
栈内存储的是高度递增的下标。对于每一个直方图高度，分两种情况。1：当栈空或者当前高度大于栈顶下标所指示的高度时，当前下标入栈。否则，2：当前栈顶出栈，并且用这个下标所指示的高度计算面积。而这个方法为什么只需要一个栈呢？因为当第二种情况时，for循环的循环下标回退，也就让下一次for循环比较当前高度与新的栈顶下标所指示的高度，注意此时的栈顶已经改变由于之前的出栈。

class Solution {
    public int largestRectangleArea(int[] heights) {
        int area = 0;  
        Stack stack = new Stack();  
        for (int i = 0; i < heights.length; i++) {  
            if (stack.empty() || heights[stack.peek()] < heights[i]) {  
                stack.push(i);  
            } else {  
                int start = stack.pop();  
                int width = stack.empty() ? i : i - stack.peek() - 1;  
                area = Math.max(area, heights[start] * width);  
                i--;  
            }  
        }  
        while (!stack.empty()) {  
            int start = stack.pop();  
            int width = stack.empty() ? heights.length : heights.length - stack.peek() - 1;  
            area = Math.max(area, heights[start] * width);        
        }  
        return area;  
    }
}