問題

對(duì)一個(gè)已經(jīng)序列A[1,...,n],如何尋找其第k小的元素捌刮？

算法

使它在O(n)的時(shí)間復(fù)雜度內(nèi)解決

code

package wentao.algorithm.ex;

import java.util.Arrays;

public class FindTheLastKMinNum {

    public static void main(String[] args) {
        int[] A = { 8, 33, 17, 51, 57, 49, 35, 11, 25, 37, 14, 3, 2, 13, 52,
                12, 6, 29, 32, 54, 5, 16, 22, 23, 7 };
        int K = 18;
        int theLastKMin = findTheLastKMinNum(A, 0, A.length - 1, K);
        System.out.println("第" + K + "小的數(shù)是：" + theLastKMin);
    }

    /**
     * 
    * 方法名: findTheLastKMinNum
    * 方法作用: 分治法求第k小值
    * 創(chuàng)建人：WENTAO Wan
    * 創(chuàng)建時(shí)間：2016年4月4日 下午11:15:39   
    * @param @param A
    * @param @param low
    * @param @param high
    * @param @param K
    * @param @return    
    * 返回值類型： int    
    * @throws
     */
    public static int findTheLastKMinNum(int[] A, int low, int high, int K) {

        // 設(shè)置閾值
        int p = high - low + 1;
        if (p < 6) {

            // 這里要注意新分配的空間 q+1造成干擾，不能直接Sort(A)
            Arrays.sort(A, low, p);
            return A[K - 1];
        } else {

            // 分為五段
            int q = p / 5;
            int remainder = p - q * 5;

            // 每段排序并把中項(xiàng)存入mid
            int[] mid = new int[q + 1];
            for (int i = 0; i < q; i++) {
                Arrays.sort(A, 5 * i, (i + 1) * 5);
                mid[i] = A[i * 5 + 2]; // low
            }

            // 除不盡5之后的分為一段并找出中項(xiàng)存入mid
            if (remainder > 0) {
                Arrays.sort(A, 5 * q, 5 * q + remainder);
                mid[q] = A[q * 5 + (remainder + 1) / 2 - 1];
            }

            // 中項(xiàng)集合的中項(xiàng)
            int mm = findTheLastKMinNum(mid, 0, q - 1, (q + 1) / 2);

            int[] A1 = new int[p];
            int[] A2 = new int[p];
            int[] A3 = new int[p];
            int lenA1 = 0, lenA2 = 0, lenA3 = 0;

            // 分別與中項(xiàng)比較舒岸，分為新的三段
            for (int i = low; i <= high; i++) {
                if (A[i] < mm) {
                    A1[lenA1++] = A[i];
                } else if (A[i] == mm) {
                    A2[lenA2++] = A[i];
                } else if (A[i] > mm) {
                    A3[lenA3++] = A[i];
                }
            }

            // 將三段的長(zhǎng)度與K比較判斷K的位置并遞歸
            int theLastKMin = 0;
            if (lenA1 >= K) {
                theLastKMin = findTheLastKMinNum(A1, 0, lenA1 - 1, K);
            } else if (lenA2 + lenA1 < K) {
                theLastKMin = findTheLastKMinNum(A3, 0, lenA3 - 1, K - lenA1
                        - lenA2);
            } else if (lenA1 + lenA2 >= K) {
                theLastKMin = mm;
            }

            return theLastKMin;
        }

    }
}