84. Largest Rectangle in Histogram

84. Largest Rectangle in Histogram

Given an array of integers?heights?representing the histogram's bar height where the width of each bar is?1, return?the area of the largest rectangle in the histogram.

key: Find each column whose first height value on the left and right side is less than that of that column

(If the right column is higher than this column, it makes no sense to count for?this column)

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1.? The main thing is to analyze the following three cases clearly:

Case 1: the currentelement heights[i] is greater than the top of the stack element heights[st.top()].? heights[i] >?heights[st.top()]
Case 2: the current?element heights[i] is equal to the top of the stack element heights[st.top()]?

? ? ? ? ? ? ? ? ? ? ? ? ? ??heights[i] =?heights[st.top()]
Case 3: the current element heights[i] is less than the top stack element heights[st.top()].

? ? ? ? ? ? ? ? ? ? ? ? ? ? heights[i] <?heights[st.top()]

2. initialization

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? heights =[0] + heights + [0]

If the array itself is in increasing order, e.g. [2,4,6,8], then after the stack is monotonically decreasing, it never goes to Case 3, so the final output is zero.

my solution:

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        stack = []
        heights =[0] + heights + [0]
        res = 0
        for i in range(len(heights)): #每一个柱子 都对应一次最大值。
            while stack and heights[i]<heights[stack[-1]]:
                mid = stack.pop()
                height = heights[mid]
                d = i - stack[-1] -1 #这里的stack[-1]是pop之后的。想清楚left，mid，right的index对应哪个
                res = max(res, height*d)
            stack.append(i)
        return res

Detailed monotonic stack：

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        # Monotonic Stack
        '''
        找每个柱子左右侧的第一个高度值小于该柱子的柱子
        单调栈：栈顶到栈底：从大到小（每插入一个新的小数值时，都要弹出先前的大数值）
        栈顶，栈顶的下一个元素，即将入栈的元素：这三个元素组成了最大面积的高度和宽度
        情况一：当前遍历的元素heights[i]大于栈顶元素的情况
        情况二：当前遍历的元素heights[i]等于栈顶元素的情况
        情况三：当前遍历的元素heights[i]小于栈顶元素的情况
        '''

        # 输入数组首尾各补上一个0（与42.接雨水不同的是，本题原首尾的两个柱子可以作为核心柱进行最大面积尝试
        heights.insert(0, 0)
        heights.append(0)
        stack = [0]
        result = 0
        for i in range(1, len(heights)):
            # 情况一
            if heights[i] > heights[stack[-1]]:
                stack.append(i)
            # 情况二
            elif heights[i] == heights[stack[-1]]:
                stack.pop()
                stack.append(i)
            # 情况三
            else:
                # 抛出所有较高的柱子
                while stack and heights[i] < heights[stack[-1]]:
                    # 栈顶就是中间的柱子，主心骨
                    mid_index = stack[-1]
                    stack.pop()
                    if stack:
                        left_index = stack[-1]
                        right_index = i
                        width = right_index - left_index - 1
                        height = heights[mid_index]
                        result = max(result, width * height)
                stack.append(i)
        return result

2 pointer:

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        size = len(heights)
        # 两个DP数列储存的均是下标index
        min_left_index = [0] * size
        min_right_index = [0] * size
        result = 0

        # 记录每个柱子的左侧第一个矮一级的柱子的下标
        min_left_index[0] = -1  # 初始化防止while死循环
        for i in range(1, size):
            # 以当前柱子为主心骨，向左迭代寻找次级柱子
            temp = i - 1
            while temp >= 0 and heights[temp] >= heights[i]:
                # 当左侧的柱子持续较高时，尝试这个高柱子自己的次级柱子（DP
                temp = min_left_index[temp]
            # 当找到左侧矮一级的目标柱子时
            min_left_index[i] = temp
        
        # 记录每个柱子的右侧第一个矮一级的柱子的下标
        min_right_index[size-1] = size  # 初始化防止while死循环
        for i in range(size-2, -1, -1):
            # 以当前柱子为主心骨，向右迭代寻找次级柱子
            temp = i + 1
            while temp < size and heights[temp] >= heights[i]:
                # 当右侧的柱子持续较高时，尝试这个高柱子自己的次级柱子（DP
                temp = min_right_index[temp]
            # 当找到右侧矮一级的目标柱子时
            min_right_index[i] = temp
        
        for i in range(size):
            area = heights[i] * (min_right_index[i] - min_left_index[i] - 1)
            result = max(area, result)
        
        return result

violent approach:

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        # 从左向右遍历：以每一根柱子为主心骨（当前轮最高的参照物），迭代直到找到左侧和右侧各第一个矮一级的柱子
        res = 0

        for i in range(len(heights)):
            left = i
            right = i
            # 向左侧遍历：寻找第一个矮一级的柱子
            for _ in range(left, -1, -1):
                if heights[left] < heights[i]:
                    break
                left -= 1
            # 向右侧遍历：寻找第一个矮一级的柱子
            for _ in range(right, len(heights)):
                if heights[right] < heights[i]:
                    break
                right += 1
                
            width = right - left - 1
            height = heights[i]
            res = max(res, width * height)

        return res

完结！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！

接下来 graph theory