微信红包（16腾讯（快排））

使用快速排序找中位数，找到以后检查一下中位数是否出现了总次数的一半以上。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Gift {
public:
	int getValue(vector<int> gifts, int n) {
		if (n <= 0)
			return -1;
		int start = 0;
		int end = n - 1;
		int middle = (end - start) >> 1;//中位数的下标
		int index;
		index = partition(gifts, start, end);
		while (index != middle) {
			if (index > middle) { //index>middle，说明中位数在左边，我们需要左调整end
				end = index - 1;
				index = partition(gifts, start, end);
			}
			else {  //index<middle，说明中位数在右边，需要右调整start
				start = index + 1;
				index = partition(gifts, start, end);

			}
		}
		return check_more_than_half(gifts, n, gifts[middle]);
	}
	int partition(vector<int>&a, int low, int high)//选第一个元素为枢轴
	{
		while (low < high)
		{
			while (low < high && a[high] >= a[low]) high--;//此时枢轴的下标为low
			swap(a[low], a[high]); //交换后枢轴的下标为high

			while (low < high && a[low] <= a[high]) low++; //此时枢轴的下标为high
			swap(a[low], a[high]);//交换后枢轴的下标为low
		}
		return low;//返回枢轴的下标

	}

	int check_more_than_half(vector<int>& gifts, const int n, int val) {
		int cnt = 0;
		for (auto i : gifts) {
			if (i == val)
				++cnt;
		}
		return (cnt != 0 && cnt > (n >> 1)) ? val : 0;
	}
};

int main() {
	Gift s;
	vector<int>a = { 1,2,3,2,2 };
	cout << s.getValue(a, 5) << endl;
	system("pause");
	return 0;
}

从大到小快速排序即可。

class Solution {
public:
	int findKthLargest(vector<int>& nums, int k) {
		int n = nums.size();
		if (n <= 0)
			return -1;
		int start = 0;
		int end = n - 1;
		int index;
		index = partition(nums, start, end);
		while (index != k - 1) {
			if (index > k - 1) { //index>k-1，说明要找的数在左边，我们需要左调整end
				end = index - 1;
				index = partition(nums, start, end);
			}
			else {  //index<k-1，说明要找的数在右边，需要右调整start
				start = index + 1;
				index = partition(nums, start, end);

			}
		}
		return nums[index];
	}
	int partition(vector<int>&a, int low, int high)//选第一个元素为枢轴
	{
		while (low < high)
		{
			while (low < high && a[high] <= a[low]) high--;//此时枢轴的下标为low
			swap(a[low], a[high]); //交换后枢轴的下标为high

			while (low < high && a[low] >= a[high]) low++; //此时枢轴的下标为high
			swap(a[low], a[high]);//交换后枢轴的下标为low
		}
		return low;//返回枢轴的下标

	}
};