Leetcode[215]-kth largest Element in an Array

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

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

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

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

Method One: Sort O (nlogn) using STL

int findKthLargest(vector<int>int k) {    sort(nums.begin(),nums.end());    return nums[nums.size()-k];}

Law two: Write your own quick sort O (nlogn)

intFindkthlargest ( vector<int>& Nums,intk) {QuickSort (nums,0, Nums.size ());returnNums[nums.size ()-K];}voidQuickSort ( vector<int>&nums,intLeftintright) {inti = left, j = right-1;if(I < J) {intPO = nums[i]; while(I < J) { while(I < J && Nums[j] >= po) j--;if(I < J)            {nums[i++] = nums[j]; } while(I < J && Nums[i] <= po) i++;if(I < J)            {nums[j--] = nums[i];        }} Nums[i] = PO;        QuickSort (Nums, left, i); QuickSort (Nums, i+1, right); }}

Method Three: Using the construction method time complexity O (KLOGN)

int findKthLargest(vector<int>int k){    make_heap(nums.begin(), nums.end());    for(auto i=0; i<k-1;i++){        pop_heap(nums.begin(), nums.end());        nums.pop_back();    }    return nums.front();}

Method Four: This method is seen by the Forum, O (N)

intFindkthlargest ( vector<int>& Nums,intK) {intI, M, N, pivot, head =0, tail = nums.size ()-1, MAXV; while(1) {m = head, n= tail; Pivot = nums[m++]; while(M <= N) {if(Nums[m] >= pivot) m++;Else if(Nums[n] < pivot) n--;Else{Swap (nums[m++], nums[n--]); }        }if(M-head = = k)returnPivotElse if(M-head < k)             {k-= (m-head);          head = m; }Else{tail = m1; Head = head+1; }    }}

Leetcode[215]-kth largest Element in an Array