刷题笔记——单调栈

保持栈的单调，只有栈为空或比栈顶大/小的元素才可入栈，否则出栈直到满足条件

In order to keep the stack monotonic, only when the stack is empty or the top element of the stack is smaller/bigger than the input one, the element will  be pushed into the stack. Otherwise, pop the elements from the stack until the criteria meets.

For an example:

lettcode 84

在这个问题中，应用了单调栈的思想，只有比栈顶大的元素才可入栈

出栈时，对出栈元素的左右边界进行计算，栈顶的左边界为0，其他元素左右边界均为栈中相邻元素的数组下标，出栈元素的值*边界长度为待求子矩形的面积

class Solution:
    def largestRectangleArea(self, heights: [int]) -> int:
        if len(heights) == 0:return 0
        max_stack = []
        maxs = heights[0]
        for i in range(len(heights)):
            if len(max_stack) == 0 or max_stack[-1][0] <= heights[i]:
                max_stack.append([heights[i], i])
            else:
                top = max_stack[-1][1]
                while len(max_stack) > 0 and max_stack[-1][0] > heights[i]:
                    last = max_stack.pop()
                    if len(max_stack) > 0:
                        maxs = max(maxs, last[0] * (top - max_stack[-1][1]))
                    else:
                        maxs = max(maxs, last[0] * (top + 1))
                max_stack.append([heights[i], i])

        top = max_stack[-1][1]
        while len(max_stack) > 0:
            last = max_stack.pop()
            if len(max_stack) > 0:
                maxs = max(maxs, last[0] * (top - max_stack[-1][1]))
            else:
                maxs = max(maxs, last[0] * (top + 1))

        return maxs