LeetCode - Maximum Frequency Stack
Implement FreqStack, a class which simulates the operation of a stack-like data structure. FreqStack has two functions: push(int x), which pushes an integer x onto the stack.
pop(), which removes and returns the most frequent element in the stack.
If there is a tie for most frequent element, the element closest to the top of the stack is removed and returned. Example 1: Input:
["FreqStack","push","push","push","push","push","push","pop","pop","pop","pop"],
[[],[5],[7],[5],[7],[4],[5],[],[],[],[]]
Output: [null,null,null,null,null,null,null,5,7,5,4]
Explanation:
After making six .push operations, the stack is [5,7,5,7,4,5] from bottom to top. Then: pop() -> returns 5, as 5 is the most frequent.
The stack becomes [5,7,5,7,4]. pop() -> returns 7, as 5 and 7 is the most frequent, but 7 is closest to the top.
The stack becomes [5,7,5,4]. pop() -> returns 5.
The stack becomes [5,7,4]. pop() -> returns 4.
The stack becomes [5,7]. Note: Calls to FreqStack.push(int x) will be such that 0 <= x <= 10^9.
It is guaranteed that FreqStack.pop() won't be called if the stack has zero elements.
The total number of FreqStack.push calls will not exceed 10000 in a single test case.
The total number of FreqStack.pop calls will not exceed 10000 in a single test case.
The total number of FreqStack.push and FreqStack.pop calls will not exceed 150000 across all test cases.
Hash map freq will count the frequence of elements.
Hash map m is a map of stack.
If element x has n frequence, we will push x n times in m[1], m[2] .. m[n]maxfreq records the maximum frequence.
push(x) will push x tom[++freq[x]]pop() will pop from the m[maxfreq]
class FreqStack {
HashMap<Integer, Integer> map;
HashMap<Integer, Stack<Integer>> freMap;
int mostFreq;
public FreqStack() {
map = new HashMap<>();
freMap = new HashMap<>();
mostFreq = 0;
}
public void push(int x) {
int freq = map.getOrDefault(x, 0)+1;
map.put(x, freq);
mostFreq = Math.max(mostFreq, freq);
if(!freMap.containsKey(freq)){
freMap.put(freq, new Stack<Integer>());
}
freMap.get(freq).push(x);
}
public int pop() {
int x = freMap.get(mostFreq).pop();
map.put(x, mostFreq-1);
if(freMap.get(mostFreq).size() == 0){
freMap.remove(mostFreq);
mostFreq --;
}
return x;
}
}
/**
* Your FreqStack object will be instantiated and called as such:
* FreqStack obj = new FreqStack();
* obj.push(x);
* int param_2 = obj.pop();
*/
LeetCode - Maximum Frequency Stack的更多相关文章
- 【LeetCode】895. Maximum Frequency Stack 解题报告(Python)
[LeetCode]895. Maximum Frequency Stack 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxueming ...
- [LeetCode] 895. Maximum Frequency Stack 最大频率栈
Implement FreqStack, a class which simulates the operation of a stack-like data structure. FreqStack ...
- LeetCode 895. Maximum Frequency Stack
题目链接:https://leetcode.com/problems/maximum-frequency-stack/ 题意:实现一种数据结构FreqStack,FreqStack需要实现两个功能: ...
- [Swift]LeetCode895. 最大频率栈 | Maximum Frequency Stack
Implement FreqStack, a class which simulates the operation of a stack-like data structure. FreqStack ...
- 最大频率栈 Maximum Frequency Stack
2018-10-06 22:01:11 问题描述: 问题求解: 为每个频率创建一个栈即可. class FreqStack { Map<Integer, Integer> map; Lis ...
- Maximum Frequency Stack
Implement FreqStack, a class which simulates the operation of a stack-like data structure. FreqStack ...
- 【LeetCode】栈 stack(共40题)
[20]Valid Parentheses (2018年11月28日,复习, ko) 给了一个字符串判断是不是合法的括号配对. 题解:直接stack class Solution { public: ...
- [LeetCode] 155. Min Stack 最小栈
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. pu ...
- Uncaught RangeError: Maximum call stack size exceeded 调试日记
异常处理汇总-前端系列 http://www.cnblogs.com/dunitian/p/4523015.html 开发道路上不是解决问题最重要,而是解决问题的过程,这个过程我们称之为~~~调试 记 ...
随机推荐
- Mac配置Hadoop最详细过程
Mac配置Hadoop最详细过程 原文链接: http://www.cnblogs.com/blog5277/p/8565575.html 原文作者: 博客园-曲高终和寡 https://www.cn ...
- spring(aop面向切面编程)
aop很早有研究过,但是最近想回顾下,顺便记录下,aop的优点有很多,实用性也很广,就好比最早在公司没有使用aop的时候没个业务层都要写try catch来捕获异常,来处理异常,甚至于记录异常或者日志 ...
- Redis入门指南之三(入门)
本节主要介绍Redis的5种数据类型,同时使用Python API来操作Redis,其中python版本为3.5, redis版本为4.0.2. redis-py 的API的使用可以分类为: (1)连 ...
- 漏洞复现——Apache HTTPD多后缀解析漏洞
漏洞原理:Apache Httpd支持一个文件拥有多个后缀,不同的后缀执行不同的命令,也就是说当我们上传的文件中只要后缀名含有php,该文件就可以被解析成php文件,利用Apache Httpd这个特 ...
- tensorflow中命名空间、变量命名的问题
1.简介 对比分析tf.Variable / tf.get_variable | tf.name_scope / tf.variable_scope的异同 2.说明 tf.Variable创建变量:t ...
- [sgu P155] Cartesian Tree
155. Cartesian Tree time limit per test: 0.25 sec. memory limit per test: 65536 KB input: standard i ...
- java多线程之yield,join,wait,sleep的区别
Java多线程之yield,join,wait,sleep的区别 Java多线程中,经常会遇到yield,join,wait和sleep方法.容易混淆他们的功能及作用.自己仔细研究了下,他们主要的区别 ...
- Linux内核分析--理解进程调度时机、跟踪分析进程调度和进程切换的过程
ID:fuchen1994 姓名:江军 作业要求: 理解Linux系统中进程调度的时机,可以在内核代码中搜索schedule()函数,看都是哪里调用了schedule(),判断我们课程内容中的总结是否 ...
- git 操作规范
分支描述 长期存在 online 主分支,负责记录上线版本的迭代,该分支代码与线上代码是完全一致的. dev 开发分支,该分支记录相对稳定的版本,所有的feature分支都从该分支创建. 多套开发环境 ...
- NPOI 操作excel之 将图片插入到指定位置;
//新建类 重写Npoi流方法 public class NpoiMemoryStream : MemoryStream { public NpoiMemoryStream() { AllowClos ...