Problem:

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.

For example,
Given height = [2,1,5,6,2,3],
return 10.

Analysis:

This problem is very very tirkcy, but elegnat!

The idea behind the solution is super powerful, you should clearly think about it.
Possible solution:
For each element in the height array.
1. go right until nums[left] < cur_height
2. then go left until nums[right] < cur_height
Then caculated the area of rectanlge of the encolsed region: cur_height * [(right-1) - (left-1) - 1]
They key point for each element is to find out the left broader and right broader.
Apparently, this method would take O(n^2). Can we do it more smartly?
iff height[cur-1]'s boundary is clear, if we still compare all elements before height[cur-1], which were already compared by height[cur-1]. There must be way to avoid those uncessary comparison. Algorithm:
Take advantage of a stack, the stack's elments are always in the "small-to-large" order.
Step 1: scan the height array from left to right.
(While) Once the current element <= top element at the stack. we pop out an element.
for (int i = 0; i < height.length; i++) {
while (!stack.isEmpty() && height[i] <= height[stack.peek()]) {
...
}
Note: this can guarantee all elements in the stack array in the ascending order. For each poped element, we can get its left border and right border, through following way.
1. the stack still have elements.
(not included)left border: the current top element in the stack, since it is the first element smaller than the poped element.
(not included)right border: the current element height[i]. Since only when "height[i] <= height[stack.peek()]", the poped element appears. The rectangle's area: (right border-1) - (left border+1) + 1 = (i - 1) - (stack.peek()+1) + 1 = (i - stack.peek() + 1) * cur_height. 2. the stack does not have elements.
(included)left border: It means all elements appear before the poped element are actually larger than the poped element, we could start from the first element of the array.
(not included)right border: The current element height[i]. Since only when "height[i] <= height[stack.peek()]", the poped element appears. The rectangle's area : (right border-1) - (left border) + 1 = (i - 1) - (0) + 1 = i * cur_height Implementation:
-------------------------------------------------------------------------------------------------
for (int i = 0; i < height.length; i++) {
while (!stack.isEmpty() && height[i] <= height[stack.peek()]) {
int cur_height = height[stack.pop()];
cur_max = stack.isEmpty() ? i*cur_height : (i-stack.peek()-1)*cur_height;
max = Math.max(max, cur_max);
}
stack.push(i);
}
------------------------------------------------------------------------------------------------- One idea should be kept in mind:
When we push a element into the stack, we pop out all elements larger than it.
Thus when we need to get left broder, we can directly get from stack.peek() after the poped operation. Since for all elements appeared after the pushed element, must take the pushed element as left border. Those poped elements would not affect them. <What a great idea!> Note: In this problem, we actually set one 0 at left side and one 0 at right side of the height array.
0 [height] 0.
For the above question, we must take care the case when all height elements were scaned, but there are still elements in the stack, thus we must use the right fake border. nums[height] = 0.

Solution:

public class Solution {
public int largestRectangleArea(int[] height) {
if (height == null || height.length == 0)
return 0;
Stack<Integer> stack = new Stack<Integer> ();
int max = 0, cur_max = 0;
for (int i = 0; i < height.length; i++) {
while (!stack.isEmpty() && height[i] <= height[stack.peek()]) {
int cur_height = height[stack.pop()];
cur_max = stack.isEmpty() ? i*cur_height : (i-stack.peek()-1)*cur_height;
max = Math.max(max, cur_max);
}
stack.push(i);
}
while (!stack.isEmpty()) {
int cur_height = height[stack.pop()];
cur_max = stack.isEmpty() ? height.length*cur_height : (height.length-stack.peek()-1)*cur_height;
max = Math.max(max, cur_max);
}
return max;
}
}

[LeetCode#84]Largest Rectangle in Histogram的更多相关文章

  1. 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 ...

  2. [LeetCode] 84. Largest Rectangle in Histogram 直方图中最大的矩形

    Given n non-negative integers representing the histogram's bar height where the width of each bar is ...

  3. LeetCode 84. Largest Rectangle in Histogram 直方图里的最大长方形

    原题 Given n non-negative integers representing the histogram's bar height where the width of each bar ...

  4. [leetcode]84. Largest Rectangle in Histogram直方图中的最大矩形

    Given n non-negative integers representing the histogram's bar height where the width of each bar is ...

  5. leetCode 84.Largest Rectangle in Histogram (最大矩形直方图) 解题思路和方法

    Given n non-negative integers representing the histogram's bar height where the width of each bar is ...

  6. [leetcode]84.Largest Rectangle in Histogram ,O(n)解法剖析

    Given n non-negative integers representing the histogram's bar height where the width of each bar is ...

  7. 【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 ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. RedHat7 SELinux

    SELinux(Security-Enhanced Linux) 是美国国家安全局(NSA)对于强制访问控制的实现,是 Linux历史上最杰出的新安全子系统.NSA是在Linux社区的帮助下开发了一种 ...

  2. Linux服务器常用性能监控命令汇总

    1.ifconfig 网卡数目.ip地址.Mac地址.MTU大小 eth0 Link encap:Ethernet HWaddr 00:0d:3a:50:12:e9 inet addr:10.0.0. ...

  3. nginx同时监听本机ipv4/ipv6端口

    修改nginx.conf配置文件 server { listen ; listen [::]:; } 0.0.0.0  表示本机所有ipv4地址,需要监听特定地址替换即可 [::]  表示本机所有ip ...

  4. 理解FTP协议

    为了防止无良网站的爬虫抓取文章,特此标识,转载请注明文章出处.LaplaceDemon/ShiJiaqi. http://www.cnblogs.com/shijiaqi1066/p/5186117. ...

  5. jquery值ajaxForm

    参考 http://www.360doc.com/content/13/1001/17/1542811_318406421.shtml

  6. C# 匿名表达式(Lambda表达式)

    匿名表达式 这次来说说Lambda表达式吧,反正也简单,我也不像其他高手那样强调来强调去,只讲一下方法: 准备条件如下: 第一,匿名表达式必须存在与之对应的委托. 只要存在相对应的委托就可以了.接下来 ...

  7. 函数还能这样玩儿~实现类似add(1)(2)(3)的函数

    人生的第一份前端工作找到了,感谢大神主子们给半路出家自学的我这么多的机会,很高兴正式踏上客观又乐趣满满的程序员之路,哇咔咔咔. ​ 分享一个准备面试时遇到的一个有趣的问题: 要求实现类似add(1)( ...

  8. struts2 Action 接收参数的三种方法

    刚学Struts2 时 大家可能遇到过很多问题,这里我讲一下Action 接收参数的三种方法,我曾经在这上面摔过一回.所以要警醒一下自己..... 第一种:Action里声明属性,样例:account ...

  9. hdoj (1162) 最小生成树

    Problem B Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Sub ...

  10. C语言运算符与表达式

    1 概论 计算机内存中的数据可以通过变量,常量来表示和存储,那么这些数据如何运算? C语言中提供了大量(34种)的运算符可以用来完成数据的算术,赋值,逻辑,关系,条件判断以及自增自减运算和基于二进制的 ...