[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];        }    }