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.

image.png

image.png

For example,
Given heights = [2,1,5,6,2,3],
return 10.

reference: https://www.youtube.com/watch?v=KkJrGxuQtYo

Solution：Stack

思路：
目的：找到每个柱的已自己为高度的左右边界。
左边界：就是此柱(的单调栈中的上一个柱)，因为stack是增长的(单调栈)
右边界：当前未知，所以push进栈

方式：
(1)遇到增长的就将其index push至stack；
(2)遇到减的时候得到此柱上一个柱的右边界：此柱。则能计算上一个柱以其自身的max = (其右边界-左边界) * 其高度

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

class Solution {
    public int largestRectangleArea(int[] heights) {
        if(heights == null || heights.length == 0) return 0;
        
        int max = 0;
        
        Deque<Integer> stack = new ArrayDeque<>();
        
        for(int cur = 0; cur < heights.length; cur++) {
            // (1) 遇到增长的就将其index push至stack；
            if(stack.isEmpty() || heights[cur] >= heights[stack.peek()]) {
                stack.push(cur);
            }
            // (2) 遇到减的时候得到此柱上一个柱的右边界：此柱。
            // 则能计算上一个柱以其自身的max = (其右边界-左边界) * 其高度
            else {
                int right_border = cur;
                int index = stack.pop();
                while(!stack.isEmpty() && heights[index] == heights[stack.peek()]) {  // 等高情况
                    index = stack.pop();
                }
                int left_border = stack.isEmpty() ? -1 : stack.peek();
                
                max = Math.max(max, (right_border - left_border - 1) * heights[index]);
                cur--; // 重新从此柱
            }
        }
        
        // 余下单调栈
        int right_border = stack.peek() + 1;
        while(!stack.isEmpty()) {
            int index = stack.pop();
            int left_border = stack.isEmpty() ? -1 : stack.peek();
            max = Math.max(max, (right_border - left_border - 1) * heights[index]);
        }
        return max;
    }
}