0
点赞
收藏
分享

微信扫一扫

Case07_最快效率求出乱序数组中第k小的元素

cnlinkchina 2022-02-07 阅读 63
package com.atzhanyuan.java;

/**
 * 最快效率求出乱序数组中第k小的元素:
 *      递归思想
 *      快速排序——双向扫描法
 *
 * @author hylstart
 * @create 2022-02-07 13:29
 */
public class Case08_最快效率求出乱序数组中第k小的元素 {
    public static void main(String[] args) {
        int[] num = {2,5,7,3,8,0,24,78};

        int k = selectK(num, 0, num.length - 1, 2);

        System.out.println(k);

    }

    static int selectK(int[] A, int p, int r, int k){
        int q = partition(A, p, r); //主元的下标
        int qk = q - p + 1; //主元是第几个元素
        if (qk == k)
            return A[q];
        else if (qk > k)
            return selectK(A, p, q - 1, k);
        else
            return selectK(A, q + 1, r, k - qk);

    }

    static int partition(int[] A, int p, int r){
        int pivot = A[p]; //确定主元的值
        int left = p + 1; //左指针
        int right = r; //右指针

        while (left <= right){
            while (left <= right && A[left] <= pivot)
                left++;
            while (left <= right && A[right] >= pivot)
                right--;
            if (left < right)
                swap(A, left, right);

        }
        swap(A, p, right);

        return right;
    }

    static void swap(int[] a, int b, int c){
        int temp = a[b];
        a[b] = a[c];
        a[c] = temp;
    }
}
举报

相关推荐

0 条评论