在整数数组中寻找第K大的数

题目描述:在一个整数数组中,根据快排的思路,找出数组中第K大的数

思路:

1. 利用快排思想,例如在49个元素中找第24大的元素,
2. 首先,进行一次快排,(将大的放到前半段,晓得放到后半段),假设得到的中轴为p
3. 判断p-low+1 == k,如果成立,直接输出a[p],(因为前半段有k-1个大于a[p]的元素,故a[p]为第k大的元素)
4. 如果p-low+1 > k,则第k大的元素在前半段,此时更新high = p-1,继续步骤2
5. 如果p-low+ < k,则第k大的元素在后半段,此时更新low = p+1,且k= k-(p=low+1),继续步骤2[时间复杂度为O(n)]

代码展示:

package com.bittech.Test;

/**
 * package:com.bittech.Test
 * Description:TODO
 * @date:2019/5/26
 * @Author:weiwei
 **/
public class Test0526 {
    public int findKth(int[] a, int n, int k) {
        return findKth(a, 0, n - 1, k);
    }

    public int findKth(int[] a, int low, int high, int k) {
        int part = partation(a, low, high);

        if (k == part - low + 1) {
            return a[part];
        } else if (k > part - low + 1) {
            return findKth(a, part + 1, high, k - part + low - 1);
        } else {
            return findKth(a, low, part - 1, k);
        }
    }

    public int partation(int[] a, int low, int high) {
        int key = a[low];
        while (low < high) {
            while (low < high && a[high] <= key) {
                high--;
                a[low] = a[high];
            }
            while (low < high && a[low] >= key) {
                low++;
                a[high] = a[low];
            }
            a[low] = key;
            return low;
        }
        return low;
    }
}