[Easy] 169. Majority Element
169. Majority Element
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 majority element always exist in the array.
Example 1:
Input: [3,2,3]
Output: 3
Example 2:
Input: [2,2,1,1,1,2,2]
Output: 2
Solution
Hash Table
Runtime: 48 ms, faster than 44.37% of C++ online submissions for Majority Element.
Memory Usage: 20 MB, less than 6.06% of C++ online submissions for Majority Element.
时间、空间复杂度O(n)
class Solution {public:int majorityElement(vector<int>& nums) {unordered_map<int, int> map;for(int i = 0; i < nums.size(); i++){map[nums[i]]++;}for(auto it = map.begin(); it != map.end(); ++it){if(it->second > nums.size()/2) return it->first;}return 0;}
};
Sorting
Runtime: 44 ms, faster than 59.04% of C++ online submissions for Majority Element.
Memory Usage: 19.6 MB, less than 6.06% of C++ online submissions for Majority Element.
class Solution {public:int majorityElement(vector<int>& nums) {nth_element(nums.begin(), nums.begin() + nums.size() / 2, nums.end());return nums[nums.size() / 2];}
};
void nth_element (RandomAccessIterator first, RandomAccessIterator nth,RandomAccessIterator last);
Rearranges the elements in the range [first,last),
in such a way that the element at the nth position is the element that would be in that position in a sorted sequence.nth_dement() 的执行会导致第 n 个元素被放置在适当的位置。
这个范围内，在第 n 个元素之前的元素都小于第 n 个元素，而且它后面的每个元素都会比它大。
由于是用来找众数，所以不必全排序。
平均时间复杂度O(n)
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