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.

在討論區找到6種方法，當然了最直接的方法是hash，但是看到tag 中有 分治、還有位操作，於是就打開評論看看大佬的代碼：

鏈接：大佬

Hash Table：思路做直接

Sorting：這個方法有點雞賊

Randomization：看似最耗時，但是最節省運行時間，直接取到major的概率有0.5，理論上最多隻要兩次就可以找到major；

Divide and Conquer：his idea is very algorithmic. However, the implementation of it requires some careful thought about the base cases of the recursion. The base case is that when the array has only one element, then it is the majority one. This solution takes 24ms.

Moore Voting Algorithm：這個方法貌似不行：{3,3,2,7,6}最後返回6而不是3；

Bit Manipulation：將每一個數字在計算機中都是01的組合，由於存在major數字，所有在每一個bit上必然有超過一半的數字同時爲0或者1，因此可以進行位操作來找出想要的major。但是運行時間有點長。

貼三個代碼：

bit：

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int major = 0, n = nums.size();
        for (int i = 0, mask = 1; i < 32; i++, mask <<= 1) {
            int bitCounts = 0;
            for (int j = 0; j < n; j++) {
                if (nums[j] & mask) bitCounts++;
                if (bitCounts > n / 2) {
                    major |= mask;
                    break;
                }
            }
        } 
        return major;
    } 
};

random:

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int n = nums.size();
        srand(unsigned(time(NULL)));
        while (true) {
            int idx = rand() % n;
            int candidate = nums[idx];
            int counts = 0; 
            for (int i = 0; i < n; i++)
                if (nums[i] == candidate)
                    counts++; 
            if (counts > n / 2) return candidate;
        }
    }
};

分治：

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 = majority(nums, left, mid);
        int rm = majority(nums, mid + 1, right);
        if (lm == rm) return lm;
        return count(nums.begin() + left, nums.begin() + right + 1, lm) > count(nums.begin() + left, nums.begin() + right + 1, rm) ? lm : rm;
    }
};