线性复杂度选出第k小元素、中位数、最小的k个元素(Java实现)

Tue Sep 5, 2017

600 Words|Read in about 2 Min
Tags: 算法  

封装成类:

package com.roc.algorithms.common;

import java.util.Arrays;

/**
 * 选出第k小元素、中位数、最小的k个元素(线性复杂度)
 *
 * @author imroc
 */
public class Select {


    //选出第k小元素,k为1~a.length
    public static int selectKthMin(int[] a, int k) {
        k--;
        int lo = 0, hi = a.length - 1;
        while (true) {
            int j = partition(a, lo, hi);
            if (j < k) {
                lo = j + 1;
            } else if (j > k) {
                hi = j - 1;
            } else {
                return a[k];
            }
        }
    }

    //选出中位数(比一半的元素小,比另一半的大)
    public static int selectMid(int[] a) {
        return selectKthMin(a, a.length / 2);
    }

    //选出k个最小元素,k为1~a.length
    public static int[] SelectKMin(int[] a, int k) {
        int lo = 0, hi = a.length - 1;
        while (true) {
            int j = partition(a, lo, hi);
            if (j < k) {
                lo = j + 1;
            } else if (j > k) {
                hi = j - 1;
            } else {
                return Arrays.copyOf(a, k);
            }
        }
    }

    private static int partition(int[] a, int lo, int hi) {
        int i = lo, j = hi + 1;
        while (true) {
            while (true) {
                i++;
                if (i == hi || a[i] > a[lo]) {
                    break;
                }
            }
            while (true) {
                j--;
                if (j == lo || a[j] <= a[lo]) {
                    break;
                }
            }
            if (i >= j) {
                break;
            }
            swap(a, i, j);
        }
        swap(a, lo, j);
        return j;
    }

    private static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

}

测试:

int[] a = { 9, 0, 6, 5, 8, 2, 1, 7, 4, 3};
System.out.println(Select.selectKthMin(a,1));//第1小元素:0
System.out.println(Select.selectMid(a));//中位数:4
System.out.println(Arrays.toString(Select.SelectKMin(a,5)));//最小的5个数:0~4

输出:
0
4
[0, 1, 2, 3, 4]

See Also

Tue Sep 5, 2017

600 Words|Read in about 2 Min
Tags: 算法