LeetCode 215. Kth Largest Element in an Array

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.

#include <vector>
#include <iostream>
#include <queue>
using namespace std;

// This method is easy but need O(n) memory space and O(n) time complexity.
int findKthUsingQueue(vector<int>& nums, int k) {
    // kth largest equals to (nums.size() - k + 1)th smallest.
    priority_queue<int, vector<int>, greater<int> > numsQueue;
    for(int i = 0; i < nums.size(); ++i) {
        numsQueue.push(nums[i]);
    }

    for(int i = 1;  i < nums.size() - k + 1; i++) {
        numsQueue.pop();
    }
    return numsQueue.top();
}
int partition(vector<int>& nums, int left, int right);

// it currently find kth-1 smallest number.
int findKth(vector<int>& nums, int first, int last, int k) {
    int pivot = partition(nums, first, last);
    if(pivot == k) return nums[pivot];
    else if(pivot < k) return findKth(nums, pivot + 1, last, k);
    else return findKth(nums, first, pivot - 1, k);
}

// O(n) time complexity and O(1) space.
int partition(vector<int>& nums, int left, int right) {
    int pivotElement = nums[left];
    int pivotIndex = left;
    for(int i = left + 1; i <= right; ++i) {
        if(nums[i] < pivotElement) {
            pivotIndex++;
            swap(nums[i], nums[pivotIndex]);
        }
    }
    swap(nums[pivotIndex], nums[left]);
    // the index p right now point to the pivot.
    return pivotIndex;
}
 int main(void) {
    vector<int> nums{4, 3, 2, 5, 8, 1};
    int res = findKthUsingQueue(nums, 2);
    int res1 = findKth(nums, 0, 5, 5);
    cout << res1 << endl;
}