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))
参考:https://juejin.cn/post/6859710267346026510
https://cloud.tencent.com/developer/article/1595001
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