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。
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. 方法一: 采用所谓的Moore voting algorithm: 每找出两个不同的element,就成对删除即count–,最终剩下的一定就是所求的...
The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the ma...[leetcode: Python]169. Majority Element 题目: Given an array of size n, find the majority element. The majority element is the element that ...
classSolution{publicintmajorityElement(int[] nums){intmajorityCount=nums.length /2;for(intnum1 : nums) {intcount=0;for(intnum2 : nums) {if(num2 == num1) { ++count; } }if(count > majorityCount) {returnnum1; } }thrownewIllegalArgumentException("The array does not contain a majority...
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. 要完成的函数: int majorityElement(vector<int>& nums) 说明: 1、这道题目给定长度为n的数组,要求输出某个元素的值...
Majority Element Given an array of sizen, 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. ...
Majority Element 题目描述: 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 ma...LeetCode 169. Majority Element Given an array of size n, find ...
public int majorityElement(int[] nums) { //require Arrays.sort(nums); //ensure return nums[nums.length/2]; } } 1. 2. 3. 4. 5. 6. 7. 8. 3、投票法 【复杂度】 时间O(N) 空间 O(1) 【思路】 设一个投票变量candidate,还有一个计数变量cnt...
Can you solve this real interview question? Majority Element - Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always ex
class Solution: def majorityElement(self, nums: List[int]) -> int: # 维护 majority ,表示众数 majority = 0 # 维护 count ,表示当前众数的个数 count = 0 # 遍历每个数 for num in nums: # 如果当前众数的个数为 0 ,则更新当前众数为 num if count == 0: majority = num if majority == ...