1publicclassSolution {2publicList<Integer> majorityElement(int[] nums) {3List<Integer> ans =newArrayList<Integer>();4if(nums.length <= 0)returnans;5Arrays.sort(nums);6inti = 0,len = nums.length,tmp = 0;7while(i
Majority Element - majority vote algorithm (Java) 1. 题目描述Description Link: https://leetcode.com/problems/majority-element/description/ Given an array of size n, find the majority element. The majority element is the element that appears mor......
classSolution:defmajorityElement(self,nums):maxCnt=0fornuminnums:ifmaxCnt==0:maxNum=nummaxCnt+=(1ifmaxNum==numelse-1)returnmaxNum Java: classSolution{publicintmajorityElement(int[]nums){IntegermaxCnt=0,maxNum=null;for(intnum:nums){if(maxCnt==0){maxNum=num;}maxCnt+=(num==maxNum)?1...
Java [Leetcode 169]Majority Element 题目描述: 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. 解题思路: 每找出...
代码(Java): 代码语言:javascript 代码运行次数:0 publicclassSolution{publicintmajorityElement(int[]nums){Arrays.sort(nums);// 先排序returnnums[nums.length/2];// 出现次数超过n/2次的元素排序后一定会出现在中间}} 他山之石: 现在让我们看看Discuss最hot的答案,我的做法并不是最快的,因为排序需要时间,...
You may assume that the array is non-empty and the majority element always exist in the array. 第一中方案如下:比较通用的一种解法,也比较笨 public class Solution { public int majorityElement(int[] num) { HashMap<Integer,Integer> map=new HashMap<>(); ...
169_求众数(Majority-Element) 这道题有 5 种方法,8 种实现,详细分析可以看花花酱的 YouTube 专栏。 文章目录 169_求众数(Majority-Element) 描述 解法一:暴力法 思路 Java 实现 Python 实现 复杂度分析 解法二:哈希表 思路 Java 实现 Python 实现 复杂度分析 解法三:排序 Java 实现 Python 实现 复杂度分....
import java.util.Hashtable; class Solution { public int majorityElement(int[] nums) { Hashtable table=new Hashtable(nums.length/2); int curr,currmax=0; int currcount=0; for(int i=0;i 更好的算法 我及其喜欢这个算法,代码简洁高效,只用了一遍遍历。不过它的核心思想我还没有参透,用这个代码一...
You may assume that the array is non-empty and the majority element always exist in the array. Solution class Solution { public int majorityElement(int[] nums) { if (nums == null || nums.length == 0) return -1; Map<Integer, Integer> map = new HashMap<>(); ...
Java 解法一: publicclassSolution {publicintmajorityElement(int[] nums) {intres = 0, cnt = 0;for(intnum : nums) {if(cnt == 0) {res = num; ++cnt;}elseif(num == res) ++cnt;else--cnt; }returnres; } } 下面这种解法利用到了位操作Bit Manipulation来解,将中位数按位来建立,从0到31...