【leetcode】Majority Element
题目概述:
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
You may assume that the array is non-empty and the majority element always exist in the array.
解题思路:
这个题解法很多,官方就给了七种
- Runtime: O(n2) — Brute force solution: Check each element if it is the majority element.
- Runtime: O(n), Space: O(n) — Hash table: Maintain a hash table of the counts of each element, then find the most common one.
- Runtime: O(n log n) — Sorting: As we know more than half of the array are elements of the same value, we can sort the array and all majority elements will be grouped into one contiguous chunk. Therefore, the middle (n/2th) element must also be the majority element.
- Average runtime: O(n), Worst case runtime: Infinity — Randomization: Randomly pick an element and check if it is the majority element. If it is not, do the random pick again until you find the majority element. As the probability to pick the majority element is greater than 1/2, the expected number of attempts is < 2.
- Runtime: O(n log n) — Divide and conquer: Divide the array into two halves, then find the majority element A in the first half and the majority element B in the second half. The global majority element must either be A or B. If A == B, then it automatically becomes the global majority element. If not, then both A and B are the candidates for the majority element, and it is suffice to check the count of occurrences for at most two candidates. The runtime complexity, T(n) = T(n/2) + 2n = O(n log n).
- Runtime: O(n) — Moore voting algorithm: We maintain a current candidate and a counter initialized to 0. As we iterate the array, we look at the current element x:
If the counter is 0, we set the current candidate to x and the counter to 1.
If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate.
After one pass, the current candidate is the majority element. Runtime complexity = O(n). - Runtime: O(n) — Bit manipulation: We would need 32 iterations, each calculating the number of 1's for the ith bit of all n numbers. Since a majority must exist, therefore, either count of 1's > count of 0's or vice versa (but can never be equal). The majority number’s ith bit must be the one bit that has the greater count.
Update (2014/12/24): Improve algorithm on the O(n log n) sorting solution: We do not need to 'Find the longest contiguous identical element' after sorting, the n/2th element is always the majority.
我用python的dict写了个,算在第二种方法里面吧:
class Solution2:
# @param num, a list of integers
# @return an integer
def majorityElement(self, num):
d = {}
l = len(num)
for i in num:
if d.has_key(i):
d[i] += 1
if d[i] > l/2:
return i
else:
d[i] = 1
if d[i] > l/2:
return i
另外借鉴了一下另一种思路:我们不断的同时移除两个不同的数,得到的最终的结果就是满足题意的数,这种思想可以延伸到出现次数大于n/k的情况(当然基于hash的方法也可以),就是同时移除k个不同的数,最后留下的结果就是满足题意的。
class Solution:
# @param num, a list of integers
# @return an integer
def majorityElement(self, num):
res = 0
c = 0
for i in num:
if c == 0:
res = i
c = 1
else:
if res == i:
c += 1
else:
c -= 1
return res
【leetcode】Majority Element的更多相关文章
- 【leetcode】Majority Element (easy)(*^__^*)
Given an array of size n, find the majority element. The majority element is the element that appear ...
- 【10_169】Majority Element
今天遇到的题都挺难的,不容易有会做的. 下面是代码,等明天看看Discuss里面有没有简单的方法~ Majority Element My Submissions Question Total Acc ...
- 【数组】Majority Element II
题目: Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The alg ...
- 【leetcode】Remove Element
题目概述: Given an array and a value, remove all instances of that value in place and return the new len ...
- 【leetcode】Remove Element (easy)
Given an array and a value, remove all instances of that value in place and return the new length. T ...
- 【leetcode】1287. Element Appearing More Than 25% In Sorted Array
题目如下: Given an integer array sorted in non-decreasing order, there is exactly one integer in the arr ...
- 【LeetCode】位运算 bit manipulation(共32题)
[78]Subsets 给了一个 distinct 的数组,返回它所有的子集. Example: Input: nums = [,,] Output: [ [], [], [], [,,], [,], ...
- 【LeetCode】分治法 divide and conquer (共17题)
链接:https://leetcode.com/tag/divide-and-conquer/ [4]Median of Two Sorted Arrays [23]Merge k Sorted Li ...
- 【LeetCode 229】Majority Element II
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorit ...
随机推荐
- 最简单的JavaScript模板引擎
在小公司待久了感觉自己的知识面很小,最近逛博客园和一些技术网站看大家在说JavaScript模版引擎的事儿,完全没有概念,网上一搜这是08年开始流行起来的...本来以为这是很高深的知识,后来在网上看到 ...
- javascript学习
代码放在E:\JS\js学习 学习中要学会多查手册 javascript基本介绍 js是用于web开发的脚本语言. 后面统称为 js 脚本语言是什么? 脚本语言不能独立使用,它和html/jsp/ph ...
- linux重定向
常用FD有3个,为0(stdin,标准输入).1(stdout,标准输出).2(stderr,标准错误输出),默认与keyboard.monitor有关: cmd > file 把 stdout ...
- 点评前端开发工具cortex安装使用方法
cortex安装方法: 安装最新版 sudo npm install -g cortex cortex config set registry http://registry.cortexjs.org ...
- Charles抓Android的数据包
1. 获得Mac OS的IP地址 2. 对Android手机设置代理,主机IP是步骤1中获得的IP,端口8888
- Property和attribute的区别[转]
Attribute和Property都可以翻译成“属性”,有的地方用Attribute表示“属性”,有的地方又在用Property,初 学者常常在这两个单词间“迷失”,甚至认为二者没有区别,是一样的. ...
- [Head First设计模式]山西面馆中的设计模式——建造者模式
系列文章 [Head First设计模式]山西面馆中的设计模式——装饰者模式 [Head First设计模式]山西面馆中的设计模式——观察者模式 引言 将学习融入生活中,是件很happy的事情,不会感 ...
- 再谈 $* 和 $@ 在 Bash 中的表现
除非特别说明,本文中出现的 Shell 均指 Bash 4.3.首先说一个基础知识:Shell 中的变量在展开成值(Parameter Expansion)之后,这个值在某些上下文(Context)中 ...
- R中的<-和=赋值符号的细致区别
<-创建的变量的作用范围可以在整个顶层环境,而=仅仅在一个局部环境. 但要<-创建的变量如果是在函数实参传递的时候创建的,其的作用范围可以在整个顶层环境,有一个前提条件:对应的形参在函数内 ...
- spark安装(实战)
sparksql+hive :http://lxw1234.com/archives/2015/06/294.htm 1,安装scala http://scala-lang.org/download/ ...