[leetcode]: 169. Majority Element

1.题目描述

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.
翻译：给一个长度为n的数组，找出其中出现次数>n/2的元素。假设一定存在

2.分析

统计了一下几种解法：
（1）找数组中位数。参考快排的partition函数，pivot=mid，就是中位数。或者暴力一点，直接排序后取中位数。
（2）多数投票算法Majority Vote Algorithm。
一次遍历，O(n)。大概思想是：元素相同->count++(赞同票),元素不同->count–(反对票)。大多数元素的个数>n/2。所以最后胜出的必然是大多数。
（3）对元素出现次数计数，选出出现次数>n/2。
（4）随机数。随机挑选一个元素，对这个元素计数判断是否>n/2次。有大于50%的概率会选中，所以还是比较巧妙的方法。

3.代码

c++
快排的partition

    int partition(vector<int>& nums, int left,int right) {
        int pivot = nums[left];
        int low = left;
        int high = right;
        while (low < high) {
            while (low<high && nums[high]>=pivot)
                --high;
            nums[low] = nums[high];
            while (low < high && nums[low] <= pivot)
                ++low;
            nums[high] = nums[low];
        }
        nums[low] = pivot;
        return low;
    }
    int majorityElement(vector<int>& nums) {
        int low = 0;
        int high = nums.size() - 1;
        int mid = (low + high) / 2;
        int pivot = partition(nums, low, high);
        while (pivot != mid) {
            if (pivot < mid)
                pivot = partition(nums, pivot + 1, high);
            else
                pivot = partition(nums, low, pivot - 1);
        }
        return nums[pivot];
    }

Majority Vote Algorithm

int majorityElement(vector<int>& nums) {
    int count = 0;
    int candidate;
    for (int n : nums) {
        if (count == 0)
        {
            ++count;
            candidate = n;
        }
        else {
            if (n == candidate)
                ++count;
            else
                --count;
        }
    }
    return candidate;
}

随机数

    int majorityElement(vector<int>& nums) {
        int n = nums.size();
        srand(unsigned(time(NULL)));
        while (true) {
            int index = rand() % n;
            int count = 0;
            for (int n : nums)
                if (n == nums[index])
                    ++count;
            if (count > n / 2)
                return nums[index];
        }
    }