[LeetCode] 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.

// 直接hash统计每个元素的出现次数。可知，必然仅有一个主要元素。// 因此，当某元素的出现次数超过半数的时候，可以直接返回。class Solution {public:    int majorityElement(vector<int>& nums) {        unordered_map<int, int> count;        const int n = nums.size();        for (auto num : nums) {            if (count[num]++ > n / 2)                return num;        }    }};

// 基本思路，由于majority element是占半数以上的元素，那么在把数组排序之后，其必然会占据数组的中间位置。// 作为优化，我们只需要局部排序即可。有点类似快排中的pivot。class Solution {public:    int majorityElement(vector<int>& nums) {        const int n = nums.size();        nth_element(nums.begin(), nums.begin() + n / 2, nums.end());        return nums[n / 2];    }};

class Solution {public:    int majorityElement(vector<int>& nums) {        return majorityElement(nums, 0, nums.size());    }private:    int majorityElement(vector<int>& nums, int lo, int hi) {        if (lo == hi) return nums[lo];        int mid = lo + (hi - lo) / 2;        int lme = majorityElement(nums, lo, mid);        int rme = majorityElement(nums, mid + 1, hi);        if (lme == rme) return lme;        int lmecnt = count(nums.begin() + lo, nums.begin() + hi + 1, lme);        int rmecnt = count(nums.begin() + lo, nums.begin() + hi + 1, rme);        return lmecnt > rmecnt ? lme : rme;    }};