［LeetCode］Kth Largest Element in an Array

解题思路：

1，借用快排分治的思想；

2，递归函数中，每次随机选取一个数作为标的mark；

3，把 比mark大的存入一个数组larger，比mark小的存入另外一个数组smaller

4，结束条件就是，所找的第k大的数， k ＝＝ larger.size() + 1 

// C++ 实现
class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        return find(nums, k);
    }

    int find(vector<int>& nums, int k){
        int n = nums.size();
        vector<int> smaller;
        vector<int> larger;
        int inx = rand()%n;

        for (auto elem: nums){
            if (elem > nums[inx]){
                larger.push_back(elem);
            }else if (elem < nums[inx]){
                smaller.push_back(elem);
            }
        }
        int sizeA = smaller.size();
        int sizeB = larger.size();
        if (sizeB < k && (n - sizeA) >= k){
            return nums[inx];
        }else{
            if (sizeB >= k){
                return find(larger, k);
            }else if (n - sizeA < k){
                return find(smaller, k-(n-sizeA));
            }
        }
    }
};

# python 实现
import random
class Solution:
    # @param {integer[]} nums
    # @param {integer} k
    # @return {integer}
    def find(self, nums, n, k):

        inx = int(random.random()*n)
        a = [ i for i in nums if i < nums[inx]]
        smaller = len(a)
        b = [ i for i in nums if i > nums[inx]]
        larger = len(b)

        if smaller < k and n - larger >= k:
            return nums[inx]
        else:
            if smaller >= k:
                return self.find(a, smaller, k)
            elif n - larger < k:
                return self.find(b, larger, k - (n-larger))

    def findKthLargest(self, nums, k):

        return self.find(nums, len(nums), len(nums)- k + 1)