LeetCode题解---Majority Element II 摩尔投票法 题目描述: Given an integer array of sizen, find all elements that appear more than⌊ n/3 ⌋times. The algorithm should run in linear time and in O(1) space. 分析: 因为要找出的是出现次数大于⌊ n/3 ⌋的元素,因此最多只可能存在两个...
intcount2=0; for(intnum : nums) { if(num == candidate1) { count1++; }elseif(num == candidate2) { count2++; }elseif(count1 ==0) { candidate1 = num; count1++; }elseif(count2 ==0) { candidate2 = num; count2++; }else{ count1--; count2--; } } count1 =0; count2...
Example 1: Example 2...Leetcode 229. Majority Element II Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 题意:从数组里找出所有出现次数大于n/3的数字,要求线性时间复杂度,O(1)...
在Majority Element题中,超过一半的数有且只会有一个,所以我们只要投票出一个数就行了。而这题中,超过n/3的数最多可能有两个,所以我们要记录出现最多的两个数。 首先获得出现次数最多的两个数 同样的两个candidate和对应的两个counter,如果遍历时,某个候选数和到当前数相等,则给相应计数器加1;如果两个计数...
229. Majority Element II Given an integer array of sizen, find all elements that appear more than⌊ n/3 ⌋times. Example 1: Input: nums = [3,2,3] Output: [3] Example 2: Input: nums = [1] Output: [1] Example 3: Input: nums = [1,2] ...
https://leetcode-cn.com/problems/majority-element-ii/ 看到这道题首先想到的是数据流模型中用到的MG算法,大概解释如下: 问题: 输入:N个元素,整数k 输出:以N/k的绝对误差估计每个元素出现的次数 算法:保存k-1个元素与其对应的计数器。对于一个新元素: ...
[LeetCode] Majority Element II 简介:Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.解题思路摩尔投票法。投票法的核心是找出两个候选众数进行投票,需要两遍遍历...
[LeetCode] Majority Element II 简介:Well, this problem becomes a little trickier since there may be more than one majority element. But, there can be at most two of them. Well, this problem becomes a little trickier since there may be more than one majority element. But, there can be ...
169. Majority Element 169. Majority Element 题目 解题思路 我的代码 优秀代码(适用范围更广) 最佳代码 题目 解题思路 **自己的方法:**通过Counter来计算出每个数字对应的数字出现的次数,遍历Counter找出出现次数最大值的数字。 其他方法: 通过构造set得出数字的列表,使用count来计算出现数字的个数,得到最大值。
输入：[1,2,3,2,2,2,5,4,2]输出：2分析问题哈希表法我们最容易想到的方法就是使用哈希表法，去统计每个数字出现的次数，即可很容易的求出众数。defmajorityElement(nums):#用字典来保存每个数字出现的次数data_count={}forxinnums:ifxindata_count:data_count[x]=data_count[x]+1else:data_...