快速排序与快速选择 (Java)

快速选择可以在nlogn的时间复杂度下得到一组无序数字的第k大，原理基于快速排序，每次以某个数字为中点分成的左右两段，左边都小于等于分割数，而右边都大于等于分割数，快速选择的求解速度要比全部排序后再取第k大快

/**
 * Created by tcgogogo on 16/8/18.
 */
class Solution {

    public void outPut(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            System.out.print(nums[i] + " ");
        }
        System.out.println();
    }

    public void swap(int[] nums, int i, int j) {
        if (i == j) {
            return;
        }
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }

    public void quickSort(int[] nums, int left, int right) {
        if (left > right) {
            return;
        }
        int i = left, j = right;
        while (i < j) {
            while (i < j && nums[j] >= nums[left]) {
                j --;
            }
            while (i < j && nums[i] <= nums[left]) {
                i ++;
            }
            if(i < j) {
                swap(nums, i, j);
            }
        }
        swap(nums, left, i);
        quickSort(nums, left, i - 1);
        quickSort(nums, i + 1, right);
    }

    public int quickSelect(int[] nums, int left, int right, int k) {
        if (left > right) {
            return 0;
        }
        int i = left, j = right;
        while (i < j) {
            while (i < j && nums[j] >= nums[left]) {
                j --;
            }
            while (i < j && nums[i] <= nums[left]) {
                i ++;
            }
            swap(nums, i, j);
        }
        swap(nums, left, i);
        if(k == i - left + 1) {
            return nums[i];
        }
        else if(k < i - left + 1) {
            return quickSelect(nums, left, i - 1, k);
        }
        else {
            return quickSelect(nums, i + 1, right, k - (i - left + 1));
        }
    }
}