还有Wirth的选择算法，它的实现比QuickSelect简单。Wirth的选择算法比QuickSelect慢，但经过一些改进，它变得更快。

更详细地说。使用Vladimir Zabrodsky的MODIFIND优化和3中位数的枢轴选择，并注意算法划分部分的最后步骤，我提出了以下算法(想象一下，命名为“LefSelect”):

#define F_SWAP(a,b) { float temp=(a);(a)=(b);(b)=temp; }

# Note: The code needs more than 2 elements to work
float lefselect(float a[], const int n, const int k) {
    int l=0, m = n-1, i=l, j=m;
    float x;

    while (l<m) {
        if( a[k] < a[i] ) F_SWAP(a[i],a[k]);
        if( a[j] < a[i] ) F_SWAP(a[i],a[j]);
        if( a[j] < a[k] ) F_SWAP(a[k],a[j]);

        x=a[k];
        while (j>k & i<k) {
            do i++; while (a[i]<x);
            do j--; while (a[j]>x);

            F_SWAP(a[i],a[j]);
        }
        i++; j--;

        if (j<k) {
            while (a[i]<x) i++;
            l=i; j=m;
        }
        if (k<i) {
            while (x<a[j]) j--;
            m=j; i=l;
        }
    }
    return a[k];
}

在我这里做的基准测试中，LefSelect比QuickSelect快20-30%。

下面是eladv建议的算法的实现(我也把随机pivot的实现放在这里):

public class Median {

    public static void main(String[] s) {

        int[] test = {4,18,20,3,7,13,5,8,2,1,15,17,25,30,16};
        System.out.println(selectK(test,8));

        /*
        int n = 100000000;
        int[] test = new int[n];
        for(int i=0; i<test.length; i++)
            test[i] = (int)(Math.random()*test.length);

        long start = System.currentTimeMillis();
        random_selectK(test, test.length/2);
        long end = System.currentTimeMillis();
        System.out.println(end - start);
        */
    }

    public static int random_selectK(int[] a, int k) {
        if(a.length <= 1)
            return a[0];

        int r = (int)(Math.random() * a.length);
        int p = a[r];

        int small = 0, equal = 0, big = 0;
        for(int i=0; i<a.length; i++) {
            if(a[i] < p) small++;
            else if(a[i] == p) equal++;
            else if(a[i] > p) big++;
        }

        if(k <= small) {
            int[] temp = new int[small];
            for(int i=0, j=0; i<a.length; i++)
                if(a[i] < p)
                    temp[j++] = a[i];
            return random_selectK(temp, k);
        }

        else if (k <= small+equal)
            return p;

        else {
            int[] temp = new int[big];
            for(int i=0, j=0; i<a.length; i++)
                if(a[i] > p)
                    temp[j++] = a[i];
            return random_selectK(temp,k-small-equal);
        }
    }

    public static int selectK(int[] a, int k) {
        if(a.length <= 5) {
            Arrays.sort(a);
            return a[k-1];
        }

        int p = median_of_medians(a);

        int small = 0, equal = 0, big = 0;
        for(int i=0; i<a.length; i++) {
            if(a[i] < p) small++;
            else if(a[i] == p) equal++;
            else if(a[i] > p) big++;
        }

        if(k <= small) {
            int[] temp = new int[small];
            for(int i=0, j=0; i<a.length; i++)
                if(a[i] < p)
                    temp[j++] = a[i];
            return selectK(temp, k);
        }

        else if (k <= small+equal)
            return p;

        else {
            int[] temp = new int[big];
            for(int i=0, j=0; i<a.length; i++)
                if(a[i] > p)
                    temp[j++] = a[i];
            return selectK(temp,k-small-equal);
        }
    }

    private static int median_of_medians(int[] a) {
        int[] b = new int[a.length/5];
        int[] temp = new int[5];
        for(int i=0; i<b.length; i++) {
            for(int j=0; j<5; j++)
                temp[j] = a[5*i + j];
            Arrays.sort(temp);
            b[i] = temp[2];
        }

        return selectK(b, b.length/2 + 1);
    }
}

    function nthMax(arr, nth = 1, maxNumber = Infinity) {
      let large = -Infinity;
      for(e of arr) {
        if(e > large && e < maxNumber ) {
          large = e;
        } else if (maxNumber == large) {
          nth++;
        }
      }
      return nth==0 ? maxNumber: nthMax(arr, nth-1, large);
    }

    let array = [11,12,12,34,23,34];

    let secondlargest = nthMax(array, 1);

    console.log("Number:", secondlargest);




   

首先，我们可以从未排序的数组中构建一个BST，它需要O(n)时间，从BST中我们可以找到O(log(n))中第k个最小的元素，它的总计数为O(n)。

转到这个链接的结尾:...........

http://www.geeksforgeeks.org/kth-smallestlargest-element-unsorted-array-set-3-worst-case-linear-time/

在那个('第k大元素数组')上快速谷歌返回这个:http://discuss.joelonsoftware.com/default.asp?interview.11.509587.17

"Make one pass through tracking the three largest values so far." 

(它是专门为3d最大)

这个答案是:

Build a heap/priority queue.  O(n)
Pop top element.  O(log n)
Pop top element.  O(log n)
Pop top element.  O(log n)

Total = O(n) + 3 O(log n) = O(n)