[LC] 84. Largest Rectangle in Histogram
Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.
Example:
Input: [2,1,5,6,2,3]
Output: 10 Solution 1: O(N^2) LTE
class Solution(object):
def largestRectangleArea(self, heights):
"""
:type heights: List[int]
:rtype: int
"""
if not heights:
return 0
res = -1
for i, height in enumerate(heights):
min_height = height
res = max(res, min_height)
for j in range(i + 1, len(heights)):
min_height = min(min_height, heights[j])
res = max(res, (j - i + 1) * min_height)
return res
Solution 2: O(N)
class Solution(object):
def largestRectangleArea(self, heights):
"""
:type heights: List[int]
:rtype: int
"""
if not heights:
return 0
res, index = 0, 0
stack = []
while index <= len(heights):
cur = 0 if index == len(heights) else heights[index]
if not stack or cur >= heights[stack[-1]]:
stack.append(index)
index += 1
else:
height = heights[stack.pop()]
# make sure left and right index are correct
right = index - 1
left = 0 if not stack else stack[-1] + 1
res = max(res, (right - left + 1) * height)
return res
[LC] 84. Largest Rectangle in Histogram的更多相关文章
- 84. Largest Rectangle in Histogram
https://www.cnblogs.com/grandyang/p/4322653.html 1.存储一个单调递增的栈 2.如果你不加一个0进去,[1]这种情况就会输出结果0,而不是1 3.单调递 ...
- LeetCode 84. Largest Rectangle in Histogram 单调栈应用
LeetCode 84. Largest Rectangle in Histogram 单调栈应用 leetcode+ 循环数组,求右边第一个大的数字 求一个数组中右边第一个比他大的数(单调栈 Lee ...
- 刷题84. Largest Rectangle in Histogram
一.题目说明 题目84. Largest Rectangle in Histogram,给定n个非负整数(每个柱子宽度为1)形成柱状图,求该图的最大面积.题目难度是Hard! 二.我的解答 这是一个 ...
- 【LeetCode】84. Largest Rectangle in Histogram 柱状图中最大的矩形(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 单调栈 日期 题目地址: https://leetc ...
- 【LeetCode】84. Largest Rectangle in Histogram
Largest Rectangle in Histogram Given n non-negative integers representing the histogram's bar height ...
- 84. Largest Rectangle in Histogram *HARD* -- 柱状图求最大面积 85. Maximal Rectangle *HARD* -- 求01矩阵中的最大矩形
1. Given n non-negative integers representing the histogram's bar height where the width of each bar ...
- 84. Largest Rectangle in Histogram *HARD* -- 求柱状图中的最大矩形面积
Given n non-negative integers representing the histogram's bar height where the width of each bar is ...
- [LeetCode#84]Largest Rectangle in Histogram
Problem: Given n non-negative integers representing the histogram's bar height where the width of ea ...
- LeetCode OJ 84. Largest Rectangle in Histogram
Given n non-negative integers representing the histogram's bar height where the width of each bar is ...
随机推荐
- Mybatis实现if trim(四)
1. 准备 请先完成Mybatis实现增删改查(二)和Mybatis实现条件查询(三)的基本内容 2. 关于多条件查询的疑问 在Mybatis实现条件查询(三)中我们实现了多条件(商品编码.商品名称. ...
- 22. docker 数据持久化 Data Volume
1 . 使用场景 在docker 容器被删除的时候 希望数据不丢失 2 . Volume 的使用 * 注意 在 mysql 的 Dockerfile 内 定义了 VOLUME ["var/ ...
- Java 二维数组,排序、切换顺序,查表法二进制十进制,这班查找、排序(冒泡、选择)、遍历,获取最大小值(4)
Java 二维数组,排序.切换顺序,查表法二进制十进制,折半查找.排序(冒泡.选择).遍历,获取最大小值(4)
- Java多线程求和
package test; import java.util.concurrent.*; import java.util.concurrent.locks.Lock; import java.uti ...
- Deep-Learning-with-Python] 文本序列中的深度学习
https://blog.csdn.net/LSG_Down/article/details/81327072 将文本数据处理成有用的数据表示 循环神经网络 使用1D卷积处理序列数据 深度学习模型可以 ...
- MySQL--启动和关闭MySQL服务
1.Windows下 启动服务 mysqld --console 或 net start mysql 关闭服务 mysqladmin -uroot shudown 或 net stop mysql 2 ...
- Cannot read property 'XXXX' of null/undifined
这个问题可能的原因有很多 1.如果你的js直接写在自执行函数或者head标签内的script里面,那么可以检查一下你的代码有没有用到页面里的节点,因为这样写的代码在页面加载完成之前就会开始执行,如果有 ...
- Tooltips
#include<windows.h> #include<Commctrl.h> #include"resource.h" #pragma comment( ...
- 吴裕雄--天生自然 PYTHON3开发学习:SMTP发送邮件
import smtplib smtpObj = smtplib.SMTP( [host [, port [, local_hostname]]] ) SMTP.sendmail(from_addr, ...
- 让debian8.8不休眠,debian设置不休眠模式,因为我的本本休眠了时间不准确了,得重新同步
第一步:sudo vi /etc/systemd/logind.conf /*最好备份下再修改*/ 把下面的参数改为ignoreHandleLidSwitch=ignore 第二步: sudo ser ...