栈-最大面积问题

84. 柱状图中最大的矩形
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。

示例:
输入: [2,1,5,6,2,3]
输出: 10


我们首先从左往右对数组进行遍历，借助单调栈求出了每根柱子的左边界，随后从右往左对数组进行遍历，借助单调栈求出了每根柱子的右边界。
class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        n = len(heights)
        left, right = [0] * n, [0] * n

        mono_stack = list()
        for i in range(n):
            while mono_stack and heights[mono_stack[-1]] >= heights[i]:
                mono_stack.pop()
            left[i] = mono_stack[-1] if mono_stack else -1
            mono_stack.append(i)
        
        mono_stack = list()
        for i in range(n - 1, -1, -1):
            while mono_stack and heights[mono_stack[-1]] >= heights[i]:
                mono_stack.pop()
            right[i] = mono_stack[-1] if mono_stack else n
            mono_stack.append(i)
        
        ans = max((right[i] - left[i] - 1) * heights[i] for i in range(n)) if n > 0 else 0
        return ans

优化：解决右边界优化问题
class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        n = len(heights)
        left, right = [0] * n, [n] * n

        mono_stack = list()
        for i in range(n):
            while mono_stack and heights[mono_stack[-1]] >= heights[i]:
                right[mono_stack[-1]] = i
                mono_stack.pop()
            left[i] = mono_stack[-1] if mono_stack else -1
            mono_stack.append(i)
        
        ans = max((right[i] - left[i] - 1) * heights[i] for i in range(n)) if n > 0 else 0
        return ans

42. 接雨水
给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。
上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。 感谢 Marcos 贡献此图。
示例:

输入: [0,1,0,2,1,0,1,3,2,1,2,1]
输出: 6

class Solution(object):
    def trap(self, height):
        n = len(height)
        # 同时从左往右和从右往左计算有效面积
        s1, s2 = 0, 0
        max1, max2 = 0, 0
        for i in range(n):
            if height[i] > max1:
                max1 = height[i]
            if height[n - i - 1] > max2:
                max2 = height[n - i - 1]
            s1 += max1
            s2 += max2
        # 积水面积 = S1 + S2 - 矩形面积 - 柱子面积;max1或者max2都表示最大的高度
        res = s1 + s2 - max1 * len(height) - sum(height)
        return res

85. 最大矩形
给定一个仅包含 0 和 1 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。
示例:

输入:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
输出: 6

class Solution:
    def maximalRectangle(self, matrix: List[List[str]]) -> int:
        if not matrix:
            return 0
        # 单调栈的应用 2
        def getLargestRectLayer(heights):
            ret = 0
            stack = []
            heights = [0] + heights + [0]
            for i in range(len(heights)):
                while stack and heights[stack[-1]] > heights[i]:
                    tmp = stack.pop()
                    ret = max(ret, (i - stack[-1] - 1) * heights[tmp])
                stack.append(i)
            return ret
        def getHeights(array, heights):
            for i in range(len(heights)):
                if array[i] == "1":
                    heights[i] += 1
                else:
                    heights[i]  = 0
            return heights
        # 对于每一层 获取heights数组即可
        ret = 0
        m = len(matrix)
        for i in range(m):
            if i == 0:
                heights = list(map(int, matrix[0]))
            else:
                heights = getHeights(matrix[i], heights)
            retLayer = getLargestRectLayer(heights)
            ret = max(ret, retLayer)
        return ret