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

  1. 【leetcode】Majority Element (easy)(*^__^*)

    Given an array of size n, find the majority element. The majority element is the element that appear ...

  2. 【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 ...

  3. 【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 ...

  4. 【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 ...

  5. 【leetcode】Remove Element (easy)

    Given an array and a value, remove all instances of that value in place and return the new length. T ...

  6. 【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 ...

  7. 【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: [ [], [], [], [,,], [,], ...

  8. 【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 ...

  9. 【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

  1. 运行tomcat8w.exe未安装指定的服务

    1.报错:指定的服务未安装. 2:解决方案有些写的不够详细,现在发表一遍详细操作手册 以上是解决方案备忘录

  2. 使用gulp-connect实现web服务器

    安装插件安装gulp-connect插件,安装命令如下 npm install --save-dev gulp-connect 定义web服务,gulpfile.js代码 var gulp = req ...

  3. Python 从零学起(纯基础) 笔记 之 迭代器、生成器和修饰器

    Python的迭代器. 生成器和修饰器 1. 迭代器是访问集合元素的一种方式,从第一个到最后,只许前进不许后退. 优点:不要求事先准备好整个迭代过程中的所有元素,仅仅在迭代到某个元素时才计算该元素,而 ...

  4. js jquery 选择器总结

    js jquery 选择器总结 一.原始JS选择器. id选择器:document.getElementById("test"); name选择器:document.getElem ...

  5. Web项目学习

    首先配好jdk,tomcat,下载eclipse,下载bootstrap模板,进行JDBC连接 创建项目 打开Eclipse,选择左上角的File->NEW->最后一个other,选择如下 ...

  6. U盘启动盘的制作--用U盘硬装Windows系统、或是重装Windows系统

    借助IT天空的优启通U盘启动盘的制作--用U盘装Windows系统.或是重装Windows系统之U盘启动盘的制作 1.==================================== 2.== ...

  7. ES5语法

    ES5新语法主要是体现在Object和.Array操作,同时涉及到JSON. Function.Date 和 String类型上. 1.Object ES5最大的特点是对象扩展很多方法. 新建对象:c ...

  8. [NHibernate]缓存(NHibernate.Caches)

    系列文章 [Nhibernate]体系结构 [NHibernate]ISessionFactory配置 [NHibernate]持久化类(Persistent Classes) [NHibernate ...

  9. 15个JavaScript本地存储技术的函数库和工具

    当构建更复杂的JavaScript应用程序运行在用户的浏览器是非常有用的,它可以在浏览器中存储信息,这样的信息可以被共享在不同的页面,浏览会话. 在最近的过去,这将有可能只被cookies文本文件保存 ...

  10. 使用excel计算指数平滑和移动平均

      指数平滑法 原数数据如下: 点击数据——数据分析 选择指数平滑 最一次平滑 由于我们选择的区域是B1:B22,第一个单元格“钢产量”,被当做标志,所以我们应该勾选标志.当我们勾选了标志后,列中的第 ...