84. Largest Rectangle in Histogram 

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class Solution {
    public int largestRectangleArea(int[] heights) {
        int[] heightsWithSentinel = new int[heights.length + 2]; // 右侧哨兵用于触发最后的计算，左侧哨兵用于简化左侧边界索引的获取，不用判断栈中元素是否为空
        for (int i = 0; i < heights.length; i++) {
            heightsWithSentinel[i + 1] = heights[i];
        }
        heights = heightsWithSentinel;

        int maxArea = 0;
        Stack<Integer> increaseStack = new Stack<>(); // 非严格单调递增栈，栈中存储的柱子索引

        for (int i = 0; i < heights.length; i++) {
            while (!increaseStack.isEmpty() && heights[i] < heights[increaseStack.peek()]) {
                int height = heights[increaseStack.pop()];
                int rightIndex = i; // exclusive
                int leftIndex = increaseStack.peek(); // exclusive
                int width = rightIndex - leftIndex - 1; // 注意此处的减 1 操作
                maxArea = Math.max(maxArea, height * width);
            }

            increaseStack.push(i); // 不要忘记将当前柱子索引入栈
        }

        return maxArea;
    }
}

References

84. Largest Rectangle in Histogram
剑指 Offer II 039. 直方图最大矩形面积