0321-2020-LEETCODE- searching for their first interview 40 K larger elements (quickSearch typical - Quick Search)

Quick Find Code Source: https: //leetcode-cn.com/problems/zui-xiao-de-kge-shu-lcof/solution/3chong-jie-fa-miao-sha-topkkuai-pai-dui-er-cha -sou /

public int[] getLeastNumbers(int[] arr, int k) {
        Arrays.sort(arr);
        return Arrays.copyOfRange(arr, 0, k);
    }
    @Test
    public void test13(){
        int[] arr = {3,2,1};
        System.out.println(Arrays.toString(getLeastNumbers(arr, 2)));
    }
    //寻找第K大的问题：
    public int[] getLeastNumbers1(int[] arr,int k) {
        if (k == 0 || arr.length == 0){
            return new int[0];
        }
        //最后一个参数表示的是我们要找的是下标是k - 1的数；
        return quickSearch(arr,0,arr.length - 1,k - 1);
    }

    private int[] quickSearch(int[] arr, int low, int high, int k) {
        //每快排切分一次，就返回一个切分好的值j，j代表左侧的值小于arr[j],右侧的值大于arr[j]
        int j = partition(arr,low,high);
        //如果返回的值刚好就是k,那就直接返回左侧的值就可以了
        if (j == k){
            return Arrays.copyOf(arr,k + 1);
        }
        //根据小标j与k的大小关系，决定继续切分左段还是右端
        return j > k ? quickSearch(arr,low,j - 1,k) : quickSearch(arr,j + 1,high,k);

    }

    private int partition(int[] arr, int low, int high) {
        //基准值，找第一个low下标的值
        int v = arr[low];
        //从low开始比
        int i = low;
        int j = high + 1;
        while (true){
            while (++i <= high && arr[i] < v);
            while (--j >= low && arr[j] > v);
            if (i >= j){
                break;
            }
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
        arr[low] = arr[j];
        arr[j] = v;
        return j;
    }