You may assume that the array is non-empty and the majority element always exist in the array. 方法1,采用一个map,存储每一个变量的数量,最后得出个数大于n/2的元素 1classSolution {2public:3intmajorityElement(vector<int> &num) {45intn=num.size();6intresult;7map<int,int>num2count;89for(...
}thrownewIllegalArgumentException("The array does not contain a majority element!"); } } Python 实现 classSolution:defmajorityElement(self, nums):""" :type nums: List[int] :rtype: int """majority_count =len(nums) //2fornum1innums: count =sum(1fornum2innumsifnum2 == num1)ifcount...
AI代码解释 publicclassSolution{publicintmajorityElement(int[]num){int major=num[0],count=1;for(int i=1;i<num.length;i++){if(count==0){count++;major=num[i];}elseif(major==num[i]){count++;}elsecount--;}returnmajor;}}
专栏/Leetcode力扣 169 | 多数元素 Majority Element Leetcode力扣 169 | 多数元素 Majority Element 2020年11月22日 13:516329浏览· 20点赞· 2评论 爱学习的饲养员 粉丝:7.0万文章:46 关注视频讲解 622:17 Leetcode力扣 1-300题视频讲解合集|手画图解版+代码【持续更新ing】 84.9万 800 视频 爱学习...
class Solution { public: int majorityElement(vector<int>& nums) { // 摩尔投票法 int candidate = nums[0]; // 候选值 for (unsigned int i = 1, count = 1; i < nums.size(); i++) { if (count == 0) candidate = nums[i]; ...
classSolution:defmajorityElement(self,nums:List[int])->int:current=0count=0foriinrange(len(nums)):ifnums[i]==nums[current]:count+=1else:count-=1ifcount==0:current=icount=1returnnums[current] 229 class Solution: def majorityElement(self, nums: List[int]) -> List[int]: ...
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,用这个代码一步步调试过,...
public class Solution { public int majorityElement(int[] nums) { //require int size=nums.length; Map<Integer,Integer> map=new HashMap<>(); //invariant for(int n:nums){ if(map.containsKey(n)){ if(map.get(n)==size/2)return n; ...
impl Solution { pub fn majority_element(nums: Vec<i32>) -> i32 { // 维护 majority ,表示众数 let mut majority = 0; // 维护 count ,表示当前众数的个数 let mut 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 { majo...
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