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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[Easy] 169. Majority Element

169. Majority Element

Solution

Hash Table

Sorting

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