代码(Python3) 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...
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. 代码:oj测试通过 Runtime: 197 ms 1classSolution:2#@param num...
1classSolution:2#@param num, a list of integers3#@return an integer4defmajorityElement(self, num):5dic ={}6foriinrange(len(num)):7if(num[i])notindic:8dic[num[i]] = 19else:10dic[num[i]] += 111l = [(a,b)fora, bindic.items()]12return(sorted(l,key =lambdax:x[1]))[-...
169 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]: current1,...
class Solution { public: int majorityElement(vector<int>& nums) { return majority(nums, 0, nums.size() - 1); } private: int majority(vector<int>& nums, int left, int right) { if (left == right) return nums[left]; int mid = left + ((right - left) >> 1); int lm = major...
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,他说他的时间复杂度为O(1),看了一下代码,他的...
You may assume that the array is non-empty and the majority element always exist in the array. 就是使用HashMap直接计数。 代码如下: import java.util.HashMap; import java.util.Map; public class Solution { public int majorityElement(int[] nums) ...
https://leetcode.com/discuss/19151/solution-computation-space-problem-can-extended-situation http://m.blog.csdn.net/blog/wenyusuran/40780253 解决方案: classSolution{public:intmajorityElement(vector<int>& nums){intsize = nums.size();intvote =0;intcount =0;for(inti =0;i < size;i++) ...
169. Majority Element刷题笔记 简单题,可以用字典完成。注意字典迭代使用的是键值key AI检测代码解析 class Solution: def majorityElement(self, nums: List[int]) -> int: dic = {} for num in nums: if num in dic: dic[num] += 1 else:...
The brute force solution does not allocate additional space proportional to the input size. Approach #2 HashMap Intuition We know that the majority element occurs more than [n/2] times, and a HashMap allows us to count element occurrences efficiently. ...