求柱状图中最大的矩形

问题：

# 给定 n 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1 。 
# 
#  求在该柱状图中，能够勾勒出来的矩形的最大面积。

方法一：暴力

# leetcode submit region begin(Prohibit modification and deletion)
class Solution(object):
    def largestRectangleArea(self, heights):
        """
        :type heights: List[int]
        :rtype: int
        """
        length = len(heights)
        max_area = 0
        for i in range(length):
            min_height = math.inf
            for j in range(i, length):
                min_height = min(min_height, heights[j])
                area = min_height * (j-i+1)
                max_area = max(area, max_area)
        return max_area
# leetcode submit region end(Prohibit modification and deletion)

方法二：优化，遍历每个柱子，左右扩展找满足条件的左柱和右柱

# leetcode submit region begin(Prohibit modification and deletion)
class Solution(object):
    def largestRectangleArea(self, heights):
        """
        :type heights: List[int]
        :rtype: int
        """
        length = len(heights)
        max_area = 0
        for i in range(length):
            current_height = heights[i]
            left = right = i
            while (left-1 >= 0 and heights[left-1] >= current_height):
                left -= 1
            while (right+1 < length and heights[right+1] >= current_height):
                right += 1
            max_area = max(max_area, (right-left+1) * current_height)
        return max_area
# leetcode submit region end(Prohibit modification and deletion)

方法三：维护一个单调栈，向左右寻找首次小于当前元素值的柱子索引

heights = [2,1,5,6,2,3]
def largestRectangeArea(heights):
    heights = [0] + heights + [0]   # 前后插入零，处理剩余情况
    stack = []
    area = 0
    for i in range(len(heights)):  # 遍历柱子，计算该索引前一个柱子可能的最大面积
        while stack and heights[stack[-1]] > heights[i]:    # 栈不空且遍历到柱子小于栈顶元素，可以计算面积了
            cur = stack.pop()
            area = max(area, heights[cur]*(i-1-stack[-1]))  # 宽度是前一个柱子到单调栈栈顶元素的距离
        stack.append(i)
    return area
print(largestRectangeArea(heights))