kth-largest-element

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.

思路：

　　1.利用快速排序partition的思想

　　2.partition函数返回pivot的下标, 位于pivot左边的数都小于等于pivot,位于pivote右边的数都大于等于pivote.

　　3.当pivote + 1 == nums.length - k + 1 时，表明 nums[pivote] 为 从大到小的第K个的数

　　  当 pivote + 1 < nums.length - k + 1  时， 表明 第k大的数在pivote 的右边

　　　当 pivote + 1 > nums.length - k + 1 时，表明第K大的书在pivote的左边

class Solution {
    /*
     * @param k : description of k
     * @param nums : array of nums
     * @return: description of return
     */
    public int kthLargestElement(int k, int[] nums) {
        // write your code here
        if (nums == null || nums.length == 0) {
            return 0;
        }
        if (k <= 0) {
            return 0;
        }
        return helper(nums, 0, nums.length - 1, nums.length - k + 1);

    }
    /*
    K 为小于等于第K大的数的个数
    */
    public int helper(int[] nums, int l, int r, int k) {
        if (l == r) {
            return nums[l];
        }
        int position = partition(nums, l, r);
        if (position + 1 == k) {
            return nums[position];
        } else if (position + 1 < k) {
            return helper(nums, position + 1, r, k);
        }  else {
            return helper(nums, l, position - 1, k);
        }
    }
    public int partition(int[] nums, int l, int r) {
        if (l == r) {
            return l;
        }
        int left = l, right = r;
        int now = nums[left];
        while (left < right) {
            while (left < right && nums[right] >= now) {
                right--;
            }
            nums[left] = nums[right];
            while (left < right && nums[left] <= now) {
                left++;
            }
            nums[right] = nums[left];
        }
        nums[left] = now;
        return left;
    }
};