Leetcode 215. Kth Largest Element in an Array

/** * How quick sort works. * First choose a pivot, say the first element of the array, * set two pointer low and high point to the start and end of the array, * from high to low, find a element that is samller than the pivot, replace nums[low] with nums[high] * then from low to high, find a element that is greater than the pivot, replace nums[high] with nums[low], * untill low == high, * set nums[low] as pivot, finish a partition. * Do the same for array [0, ..., index_of_pivot-1] and [index_of_pivot+1, ..., nums.length-1]. * Therefore, index_of_pivot is the index_of_pivot largetest element of the array. * For this problem. * So, if k > index_of_pivot, we need to move to the left, * else if k < index_of_pivot, we need to move to the right, * else index_of_pivot is k.  */ public class Solution {    public int findKthLargest(int[] nums, int k) {        // note that pos is the (pos+1) samllest element in the array        int low = 0, high = nums.length-1, pos = nums.length - k;                // k is always valid        while (true) {            int idx = partition(nums, low, high);            if (idx > pos) high = idx - 1;            else if (idx < pos) low = idx + 1;            else return nums[idx];        }    }        // returns (low+1)th samllest element in the array    public static int partition(int[] nums, int low, int high) {        int pivot = nums[low];        while (low < high) {            // from high to low, find an elem that is less than the pivot            while (low < high && nums[high] >= pivot) --high;            nums[low] = nums[high];            // from low to high, find an elem that is greater than the pivot            while (low < high && nums[low] <= pivot) ++low;            nums[high] = nums[low];        }        nums[low] = pivot;        return low;    }}

AVG: T(n) = T(n/2) + O(n) = O(n)
WORST: T(n) = T(n-1) + O(n) = O(n^2)