数组中的第K个最大元素--leecode 分治算法求解

public static int findKthLargest(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        //将第K个最大元素转换为第l个最小元素
        int l = nums.length + 1 - k;
        return findKthLargestWithIndex(nums, 0, nums.length - 1, l);
    }

    public static int findKthLargestWithIndex(int[] nums, int left, int right, int l) {
        int element = 0;
        int p = SortAlgorithm.randomPartitionEx(nums, left, right);
        if (p - left + 1 == l) {
            element = nums[p];
        }
        if (p - left + 1 > l) {
            element = findKthLargestWithIndex(nums, left, p, l);
        }
        if (p - left + 1 < l) {
            element = findKthLargestWithIndex(nums, p + 1, right, l - (p - left + 1));
        }
        return element;
    }