LeetCode: 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.

本来的想法就是最naive的想法。对应每一个bar，在它的左右两侧延伸，直到遇到一个比这个值小的bar，计算面积。

可以通过，就是时间比较长。

后来在网上看到一种O(n)的算法。。我发现很多我做出来的题都不是最优的算法，都是一些naive的算法。

Actually, we can decrease the complexity by using stack to keep track of the height and start indexes. Compare the current height with previous one.

Case 1: current > previous (top of height stack)
Push current height and index as candidate rectangle start position.

Case 2: current = previous
Ignore.

Case 3: current < previous
Need keep popping out previous heights, and compute the candidate rectangle with height and width (current index - previous index). Push the height and index to stacks.

每一次弹出来一个值的时候，用这个值乘上它与当前值横坐标的差，就可以计算出以被弹出的值为当前值，向两侧延伸，所得到的面积。

这是因为在入栈的时候，之前的肯定是要比他小的。现在又遇到一个比他小的，所以这个范围一定是他的最大范围。

最后所有元素处理之后，还有没弹出栈的，说明这个值在他后面一直没有遇到比他小的。所以它可以一直延伸到数组的末尾。

public int largestRectangleArea2(int[] height) {
        if (height == null || height.length == 0) return 0;
        
        int maxArea = 0;
        Stack<Integer> stackHeight = new Stack<Integer>();
        Stack<Integer> stackIndex = new Stack<Integer>();
        
        for (int i = 0; i < height.length; i++) {
            //case 1
            if (stackHeight.isEmpty() || height[i] > stackHeight.peek()) {
                stackHeight.push(height[i]);
                stackIndex.push(i);
            } else if (height[i] < stackHeight.peek()) {
                // case 3
                int lastIndex = 0;
                while (stackHeight.isEmpty() == false && height[i] < stackHeight.peek()) {
                    lastIndex = stackIndex.pop();
                    int tempArea = stackHeight.pop() * (i - lastIndex);
                    if (maxArea < tempArea) {
                        maxArea = tempArea;
                    }
                }
                stackHeight.push(height[i]);
                stackIndex.push(lastIndex);
            }
        }
        
        while(stackHeight.isEmpty() == false) {
            int tempArea = stackHeight.pop() * (height.length - stackIndex.pop());
            if (tempArea > maxArea) {
                maxArea = tempArea;
            }
        }
        return maxArea;
    }