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.

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

Subscribe to see which companies asked this question.

这题我直接想到的就是快速排序，快速排序已经是用了分治算法来实现较快的排序了，

以下是快速排序

void Qsort(int a[], int low, int high)
{
    if(low >= high)
    {
        return;
    }
    int first = low;
    int last = high;
    int key = a[first];
  while(first < last)
    {
        while(first < last && a[last] >= key)
        {
            --last;
        }
        a[first] = a[last];
        while(first < last && a[first] <= key)
        {
            ++first;
        }        
       a[last] = a[first];    
    }
    a[first] = key;
    Qsort(a, low, first-1);
    Qsort(a, first+1, high);
}

但是在看了网站给出的答案后，发现了有更快的算法

class Solution {
public:
int partition(vector<int>& nums, int left, int right) {  
    int pivot = nums[left];  
    int l = left + 1, r = right;  
    while (l <= r) {  
        if (nums[l] < pivot && nums[r] > pivot)  
            swap(nums[l++], nums[r--]);  
        else if (nums[l] >= pivot) l++;  
        else if (nums[r] <= pivot) r--;  
    }  
    swap(nums[left], nums[r]);  
    return r;  
}  
int findKthLargest(vector<int>& nums, int k) {  
    int left = 0, right = nums.size() - 1;  
    while (true) {  
        int pos = partition(nums, left, right);  
        if (pos == k - 1) return nums[pos];  
        if (pos < k - 1) left = pos + 1;  
        else right = pos - 1;  
    }  
}  
};  

1. Initialize left to be 0 and right to be nums.size() - 1;
2. Partition the array, if the pivot is at the k-1-th position, return it (we are done);
3. If the pivot is right to the k-1-th position, update right to be the left neighbor of the pivot;
4. Else update left to be the right neighbor of the pivot.
5. Repeat 2.

但是对其想法的由来却不是很理解。。。希望有大神解答