47

这是一个面试问题:给定一个整数数组,找到最大值。和分钟。使用最小比较。

显然,我可以循环两次数组并~2n在最坏的情况下使用比较,但我想做得更好。

4

14 回答 14

79
1. Pick 2 elements(a, b), compare them. (say a > b)
2. Update min by comparing (min, b)
3. Update max by comparing (max, a)

这样,您将对 2 个元素进行 3 次比较,相当于元素的3N/2总比较N

于 2012-11-24T19:07:41.170 回答
15

试图通过 srbh.kmr 改进答案。假设我们有序列:

A = [a1, a2, a3, a4, a5]

比较a1&a2并计算min12, max12:

if (a1 > a2)
  min12 = a2
  max12 = a1
else
  min12 = a1
  max12 = a2

同样计算min34, max34。既然a5是一个人,就保持原样吧……

现在比较min12&min34和计算min14,同样计算max14。最后比较min14&a5来计算min15。同样计算max15

总共只有6个比较!

该解决方案可以扩展到任意长度的数组。可能可以通过类似于合并排序的方法来实现(将数组分成两半并计算min max每一半)。

更新:这是 C 中的递归代码:

#include <stdio.h>

void minmax (int* a, int i, int j, int* min, int* max) {
  int lmin, lmax, rmin, rmax, mid;
  if (i == j) {
    *min = a[i];
    *max = a[j];
  } else if (j == i + 1) {
    if (a[i] > a[j]) {
      *min = a[j];
      *max = a[i];
    } else {
      *min = a[i];
      *max = a[j];
    }
  } else {
    mid = (i + j) / 2;
    minmax(a, i, mid, &lmin, &lmax);
    minmax(a, mid + 1, j, &rmin, &rmax);
    *min = (lmin > rmin) ? rmin : lmin;
    *max = (lmax > rmax) ? lmax : rmax;
  }
}

void main () {
  int a [] = {3, 4, 2, 6, 8, 1, 9, 12, 15, 11};
  int min, max;
  minmax (a, 0, 9, &min, &max);
  printf ("Min : %d, Max: %d\n", min, max);
}

现在我无法根据N(数组中的元素数量)确定确切的比较次数。但是很难看出一个人怎么能低于这么多的比较。

更新:我们可以计算出如下比较的数量:

在这棵计算树的底部,我们从原始数组中形成整数对。所以我们有N / 2叶子节点。对于这些叶节点中的每一个,我们只进行 1 次比较。

通过参考完美二叉树的属性,我们有:

leaf nodes (L) = N / 2 // known
total nodes (n) = 2L - 1 = N - 1
internal nodes = n - L = N / 2 - 1

对于每个内部节点,我们进行 2 次比较。因此,我们有N - 2比较。除了N / 2叶节点的比较之外,我们还有(3N / 2) - 2总比较。

所以,这可能是他的答案中暗示的解决方案 srbh.kmr。

于 2012-11-24T22:59:08.760 回答
5

分而治之!

1,3,2,5

对于这个发现 min,max 将进行 6 次比较

但把它们分开

1,3 ---> will give min 1 and max 3 in one comparison
2,5 ---> will give min 2 and max 5 in one comparison 

现在我们可以比较两个最小值和最大值

min(1,2) --> will give the final min as 1 (one comparison)
max(3,5) ---> will give the final max as 5 (one comparison)

所以总共有四个比较来找到最小值和最大值。

于 2014-06-11T06:07:44.873 回答
5

一种稍微不同的方法,它使用整数算术而不是比较(没有明确禁止)

for(int i=0;i<N;i++) {
  xmin += x[i]-xmin & x[i]-xmin>>31;
  xmax += x[i]-xmax & xmax-x[i]>>31;
}
于 2016-01-07T14:59:41.817 回答
4

蛮力更快!

我希望有人能告诉我我的方式的错误,在这里,但是,......</p>

我比较了蛮力方法与(更漂亮的)递归分而治之的实际运行时间。典型结果(每个函数调用 10,000,000 次):

Brute force :
  0.657 seconds 10 values => 16 comparisons.  Min @ 8, Max @ 10
  0.604 seconds 1000000 values => 1999985 comparisons.  Min @ 983277, Max @ 794659
Recursive :
  1.879 seconds 10 values => 13 comparisons.  Min @ 8, Max @ 10
  2.041 seconds 1000000 values => 1499998 comparisons.  Min @ 983277, Max @ 794659

令人惊讶的是,对于 10 个项目的数组,蛮力方法的速度大约快 2.9 倍,而对于 1,000,000 个项目的数组,则快 3.4 倍。

显然,比较的次数不是问题,而可能是重新分配的次数,以及调用递归函数的开销(我不知道为什么 1,000,000 个值比 10 个值运行得快,但确实如此!)。

警告:我在 VBA 中而不是 C 中执行此操作,并且我正在比较双精度数字并将索引返回到 Min 和 Max 值的数组中。

这是我使用的代码(此处不包括类 cPerformanceCounter,但使用 QueryPerformanceCounter 进行高分辨率计时):

Option Explicit

'2021-02-17

Private Const MIN_LONG As Long = -2147483648#

Private m_l_NumberOfComparisons As Long

Sub Time_MinMax()

   Const LBOUND_VALUES As Long = 1

   Dim l_pcOverall As cPerformanceCounter
   Dim l_d_Values() As Double
   Dim i As Long, _
       k As Long, _
       l_l_UBoundValues As Long, _
       l_l_NumberOfIterations As Long, _
       l_l_IndexOfMin As Long, _
       l_l_IndexOfMax As Long

   Set l_pcOverall = New cPerformanceCounter

   For k = 1 To 2

      l_l_UBoundValues = IIf(k = 1, 10, 1000000)

      ReDim l_d_Values(LBOUND_VALUES To l_l_UBoundValues)

      'Assign random values
      Randomize '1 '1 => the same random values to be used each time
      For i = LBOUND_VALUES To l_l_UBoundValues
         l_d_Values(i) = Rnd
      Next i
      For i = LBOUND_VALUES To l_l_UBoundValues
         l_d_Values(i) = Rnd
      Next i

      'This keeps the total run time in the one-second neighborhood
      l_l_NumberOfIterations = 10000000 / l_l_UBoundValues

      '——————— Time Brute Force Method —————————————————————————————————————————
      l_pcOverall.RestartTimer

      For i = 1 To l_l_NumberOfIterations

         m_l_NumberOfComparisons = 0

         IndexOfMinAndMaxDoubleBruteForce _
               l_d_Values, _
               LBOUND_VALUES, _
               l_l_UBoundValues, _
               l_l_IndexOfMin, _
               l_l_IndexOfMax

      Next

      l_pcOverall.ElapsedSecondsDebugPrint _
            3.3, , _
            " seconds Brute-Force " & l_l_UBoundValues & " values => " _
            & m_l_NumberOfComparisons & " comparisons. " _
            & " Min @ " & l_l_IndexOfMin _
            & ", Max @ " & l_l_IndexOfMax, _
            True
      '——————— End Time Brute Force Method —————————————————————————————————————

      '——————— Time Brute Force Using Individual Calls —————————————————————————
      l_pcOverall.RestartTimer

      For i = 1 To l_l_NumberOfIterations

         m_l_NumberOfComparisons = 0

         l_l_IndexOfMin = IndexOfMinDouble(l_d_Values)
         l_l_IndexOfMax = IndexOfMaxDouble(l_d_Values)

      Next

      l_pcOverall.ElapsedSecondsDebugPrint _
            3.3, , _
            " seconds Individual  " & l_l_UBoundValues & " values => " _
            & m_l_NumberOfComparisons & " comparisons. " _
            & " Min @ " & l_l_IndexOfMin _
            & ", Max @ " & l_l_IndexOfMax, _
            True
      '——————— End Time Brute Force Using Individual Calls —————————————————————

      '——————— Time Recursive Divide and Conquer Method ————————————————————————
      l_pcOverall.RestartTimer

      For i = 1 To l_l_NumberOfIterations

         m_l_NumberOfComparisons = 0

         IndexOfMinAndMaxDoubleRecursiveDivideAndConquer _
               l_d_Values, _
               LBOUND_VALUES, _
               l_l_UBoundValues, _
               l_l_IndexOfMin, l_l_IndexOfMax

      Next

      l_pcOverall.ElapsedSecondsDebugPrint _
            3.3, , _
            " seconds Recursive   " & l_l_UBoundValues & " values => " _
            & m_l_NumberOfComparisons & " comparisons. " _
            & " Min @ " & l_l_IndexOfMin _
            & ", Max @ " & l_l_IndexOfMax, _
            True
      '——————— End Time Recursive Divide and Conquer Method ————————————————————

   Next k

End Sub

'Recursive divide and conquer
Sub IndexOfMinAndMaxDoubleRecursiveDivideAndConquer( _
      i_dArray() As Double, _
      i_l_LBound As Long, _
      i_l_UBound As Long, _
      o_l_IndexOfMin As Long, _
      o_l_IndexOfMax As Long)

   Dim l_l_IndexOfLeftMin As Long, _
       l_l_IndexOfLeftMax As Long, _
       l_l_IndexOfRightMin As Long, _
       l_l_IndexOfRightMax As Long, _
       l_l_IndexOfMidPoint As Long

   If (i_l_LBound = i_l_UBound) Then 'Only one element

      o_l_IndexOfMin = i_l_LBound
      o_l_IndexOfMax = i_l_LBound

   ElseIf (i_l_UBound = (i_l_LBound + 1)) Then 'Only two elements

      If (i_dArray(i_l_LBound) > i_dArray(i_l_UBound)) Then
         o_l_IndexOfMin = i_l_UBound
         o_l_IndexOfMax = i_l_LBound
      Else
         o_l_IndexOfMin = i_l_LBound
         o_l_IndexOfMax = i_l_UBound
      End If

      m_l_NumberOfComparisons = m_l_NumberOfComparisons + 1

   Else 'More than two elements => recurse

      l_l_IndexOfMidPoint = (i_l_LBound + i_l_UBound) / 2

      'Find the min of the elements in the left half
      IndexOfMinAndMaxDoubleRecursiveDivideAndConquer _
            i_dArray, _
            i_l_LBound, _
            l_l_IndexOfMidPoint, _
            l_l_IndexOfLeftMin, _
            l_l_IndexOfLeftMax

      'Find the min of the elements in the right half
      IndexOfMinAndMaxDoubleRecursiveDivideAndConquer i_dArray, _
            l_l_IndexOfMidPoint + 1, _
            i_l_UBound, _
            l_l_IndexOfRightMin, _
            l_l_IndexOfRightMax

      'Return the index of the lower of the two values returned
      If (i_dArray(l_l_IndexOfLeftMin) > i_dArray(l_l_IndexOfRightMin)) Then
         o_l_IndexOfMin = l_l_IndexOfRightMin
      Else
         o_l_IndexOfMin = l_l_IndexOfLeftMin
      End If

      m_l_NumberOfComparisons = m_l_NumberOfComparisons + 1

      'Return the index of the lower of the two values returned
      If (i_dArray(l_l_IndexOfLeftMax) > i_dArray(l_l_IndexOfRightMax)) Then
         o_l_IndexOfMax = l_l_IndexOfLeftMax
      Else
         o_l_IndexOfMax = l_l_IndexOfRightMax
      End If

      m_l_NumberOfComparisons = m_l_NumberOfComparisons + 1

   End If

End Sub

Sub IndexOfMinAndMaxDoubleBruteForce( _
      i_dArray() As Double, _
      i_l_LBound As Long, _
      i_l_UBound As Long, _
      o_l_IndexOfMin As Long, _
      o_l_IndexOfMax As Long)

   Dim i As Long

   o_l_IndexOfMin = i_l_LBound
   o_l_IndexOfMax = o_l_IndexOfMin

   For i = i_l_LBound + 1 To i_l_UBound

      'Usually we will do two comparisons
      m_l_NumberOfComparisons = m_l_NumberOfComparisons + 2

      If (i_dArray(i) < i_dArray(o_l_IndexOfMin)) Then

         o_l_IndexOfMin = i

         'We don't need to do the ElseIf comparison
         m_l_NumberOfComparisons = m_l_NumberOfComparisons - 1

      ElseIf (i_dArray(i) > i_dArray(o_l_IndexOfMax)) Then

         o_l_IndexOfMax = i

      End If
   Next i

End Sub

Function IndexOfMinDouble( _
      i_dArray() As Double _
      ) As Long

   Dim i As Long

   On Error GoTo EWE

   IndexOfMinDouble = LBound(i_dArray)

   For i = IndexOfMinDouble + 1 To UBound(i_dArray)

      If (i_dArray(i) < i_dArray(IndexOfMinDouble)) Then
         IndexOfMinDouble = i
      End If

      m_l_NumberOfComparisons = m_l_NumberOfComparisons + 1

   Next i

   On Error GoTo 0
   Exit Function
EWE:
   On Error GoTo 0
   IndexOfMinDouble = MIN_LONG
End Function

Function IndexOfMaxDouble( _
      i_dArray() As Double _
      ) As Long

   Dim i As Long

   On Error GoTo EWE

   IndexOfMaxDouble = LBound(i_dArray)

   For i = IndexOfMaxDouble + 1 To UBound(i_dArray)

      If (i_dArray(i) > i_dArray(IndexOfMaxDouble)) Then
         IndexOfMaxDouble = i
      End If

      m_l_NumberOfComparisons = m_l_NumberOfComparisons + 1

   Next i

   On Error GoTo 0
   Exit Function
EWE:
   On Error GoTo 0
   IndexOfMaxDouble = MIN_LONG
End Function
于 2014-07-01T16:22:12.143 回答
4

阅读问题和答案后,我决定尝试几个版本(在 C# 中)。
我认为最快的是 Anton Knyazyev 的(无分支),它不是(在我的盒子上)。
结果:

/*                comp.  time(ns)
      minmax0     3n/2    855
      minmax1     2n      805
      minmax2     2n     1315 
      minmax3     2n      685          */

为什么 minmax1 和 minmax3 更快?可能是因为“分支预测器”做得很好,每次迭代发现新的最小值(或最大值)的机会都会减少,因此预测会变得更好。
总而言之,这是一个简单的测试。我确实意识到我的结论可能是:-
为时过早。
- 不适用于不同的平台。
假设它们是指示性的。
编辑:盈亏平衡点 minmax0,minmax3:~100 个项目,
10,000 个项目:minmax3 比 minmax0 快 ~3.5 倍。

using System;
using sw = System.Diagnostics.Stopwatch;
class Program
{
    static void Main()
    {
        int n = 1000;
        int[] a = buildA(n);
        sw sw = new sw();
        sw.Start();
        for (int i = 1000000; i > 0; i--) minMax3(a);
        sw.Stop();
        Console.Write(sw.ElapsedMilliseconds);
        Console.Read();
    }

    static int[] minMax0(int[] a)  // ~3j/2 comp.
    {
        int j = a.Length - 1;
        if (j < 2) return j < 0 ? null :
            j < 1 ? new int[] { a[0], a[0] } :
            a[0] < a[1] ? new int[] { a[0], a[1] } :
                          new int[] { a[1], a[0] };
        int a0 = a[0], a1 = a[1], ai = a0;
        if (a1 < a0) { a0 = a1; a1 = ai; }
        int i = 2;
        for (int aj; i < j; i += 2)
        {
            if ((ai = a[i]) < (aj = a[i + 1]))  // hard to predict
            { if (ai < a0) a0 = ai; if (aj > a1) a1 = aj; }
            else
            { if (aj < a0) a0 = aj; if (ai > a1) a1 = ai; }
        }
        if (i <= j)
        { if ((ai = a[i]) < a0) a0 = ai; else if (ai > a1) a1 = ai; }
        return new int[] { a0, a1 };
    }

    static int[] minMax1(int[] a)  // ~2j comp.  
    {
        int j = a.Length;
        if (j < 3) return j < 1 ? null :
            j < 2 ? new int[] { a[0], a[0] } :
            a[0] < a[1] ? new int[] { a[0], a[1] } :
                          new int[] { a[1], a[0] };
        int a0 = a[0], a1 = a0, ai = a0;
        for (int i = 1; i < j; i++)
        {
            if ((ai = a[i]) < a0) a0 = ai;
            else if (ai > a1) a1 = ai;
        }
        return new int[] { a0, a1 };
    }

    static int[] minMax2(int[] a)  // ~2j comp.  
    {
        int j = a.Length;
        if (j < 2) return j == 0 ? null : new int[] { a[0], a[0] };
        int a0 = a[0], a1 = a0;
        for (int i = 1, ai = a[1], aj = ai; ; aj = ai = a[i])
        {
            ai -= a0; a0 += ai & ai >> 31;
            aj -= a1; a1 += aj & -aj >> 31;
            i++; if (i >= j) break;
        }
        return new int[] { a0, a1 };
    }

    static int[] minMax3(int[] a)  // ~2j comp.
    {
        int j = a.Length - 1;
        if (j < 2) return j < 0 ? null :
            j < 1 ? new int[] { a[0], a[0] } :
            a[0] < a[1] ? new int[] { a[0], a[1] } :
                          new int[] { a[1], a[0] };
        int a0 = a[0], a1 = a[1], ai = a0;
        if (a1 < a0) { a0 = a1; a1 = ai; }
        int i = 2;
        for (j -= 2; i < j; i += 3)
        {
            ai = a[i + 0]; if (ai < a0) a0 = ai; if (ai > a1) a1 = ai;
            ai = a[i + 1]; if (ai < a0) a0 = ai; if (ai > a1) a1 = ai;
            ai = a[i + 2]; if (ai < a0) a0 = ai; if (ai > a1) a1 = ai;
        }
        for (j += 2; i <= j; i++)
        { if ((ai = a[i]) < a0) a0 = ai; else if (ai > a1) a1 = ai; }
        return new int[] { a0, a1 };
    }

    static int[] buildA(int n)
    {
        int[] a = new int[n--]; Random rand = new Random(0);
        for (int j = n; n >= 0; n--) a[n] = rand.Next(-1 * j, 1 * j);
        return a;
    }
}
于 2016-11-13T11:34:55.123 回答
2

递归算法的简单伪代码:

Function MAXMIN (A, low, high)
    if (high − low + 1 = 2) then 
      if (A[low] < A[high]) then
         max = A[high]; min = A[low].
         return((max, min)).
      else
         max = A[low]; min = A[high].
         return((max, min)).
      end if
   else
      mid = low+high/2
      (max_l , min_l ) = MAXMIN(A, low, mid).
      (max_r , min_r ) =MAXMIN(A, mid + 1, high).
   end if

   Set max to the larger of max_l and max_r ; 
   likewise, set min to the smaller of min_l and min_r .

   return((max, min)).
于 2014-11-16T06:06:24.653 回答
1
import java.util.*;
class Maxmin
{
    public static void main(String args[])
    {
        int[] arr = new int[10];
        Scanner in = new Scanner(System.in);
        int i, min=0, max=0;
        for(i=0; i<=9; i++)
        {
            System.out.print("Enter any number: ");
            arr[i] = in.nextInt();          
        }
        min = arr[0];
        for(i=0; i<=9; i++)
        {
            if(arr[i] > max)
            {
                max = arr[i];
            }
            if(arr[i] < min)
            {
                min = arr[i];
            }
        }
        System.out.println("Maximum is: " + max);
        System.out.println("Minimum is: " + min);
    }
}
于 2016-01-07T07:25:33.343 回答
1

到目前为止,我使用 java 的分而治之方法:

        public class code {    
    static int[] A = {444,9,8,6,199,3,0,5,3,200};
    static int min = A[0], max = A[1];
    static int count = 0;

    public void minMax(int[] A, int i, int j) {     
        if(i==j) {
            count = count + 2;
            min = Math.min(min, A[i]);
            max = Math.max(max, A[i]);
        }

        else if(j == i+1) {
            if(A[i] > A[j]) {
                count = count + 3;
                min = Math.min(min, A[j]);
                max = Math.max(max, A[i]);
            }
            else {
                count = count + 3;
                min = Math.min(min, A[i]);
                max = Math.max(max, A[j]);
            }
        }

        else {
            minMax(A,i,(i+j)/2);
            minMax(A,(i+j)/2+1,j);
        }
    }

    public static void main(String[] args) {
        code c = new code();
        if(Math.min(A[0], A[1]) == A[0]) {
            count++;
            min = A[0];
            max = A[1];
        }
        else {
            count++;
            min = A[1];
            max = A[0];
        }
        c.minMax(A,2,A.length-1);
        System.out.println("Min: "+min+" Max: "+max);
        System.out.println("Total comparisons: " + count);
    }
}
于 2016-10-31T08:57:07.300 回答
1
public static int[] minMax(int[] array){
    int [] empty = {-1,-1};
    if(array==null || array.length==0){
        return empty;
    }

    int lo =0, hi = array.length-1;
    return minMax(array,lo, hi); 

}

private static int[] minMax(int []array, int lo, int hi){

    if(lo==hi){
        int [] result = {array[lo], array[hi]}; 
        return result;
    }else if(lo+1==hi){
        int [] result = new int[2];
        result[0] = Math.min(array[lo], array[hi]);
        result[1] = Math.max(array[lo], array[hi]);
        return result;
    }else{
        int mid = lo+(hi-lo)/2;          
        int [] left = minMax(array, lo, mid);
        int [] right = minMax(array, mid+1, hi);
        int []result = new int[2];          
        result[0] = Math.min(left[0], right[0]);
        result[1] = Math.max(left[1], right[1]);             
        return result;
    }

}

public static void main(String[] args) {

    int []array = {1,2,3,4,100};
    System.out.println("min and max values are "+Arrays.toString(minMax(array)));
}
于 2017-01-05T23:47:21.003 回答
1
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    set<int> t;
    for(int i=0;i<n;i++)
    {

        int x;
        cin>>x;
        t.insert(x);
    }
    set<int>::iterator s,b;
    s=t.begin();
    b=--t.end();
    cout<< *s<<" "<<*b<<endl;


    enter code here

    return 0;
}

// 这可以在 log(n) 复杂度中完成!!!

于 2017-05-11T05:05:01.090 回答
1
if (numbers.Length <= 0)
{
    Console.WriteLine("There are no elements");
    return;
}

if (numbers.Length == 1)
{
    Console.WriteLine($"There is only one element. So min and max of this 
                        array is: {numbers[0]}");
    return;
}

if (numbers.Length == 2)
{
    if (numbers[0] > numbers[1])
    {
        Console.WriteLine($"min = {numbers[1]}, max = {numbers[0]}");
        return;
    }

    Console.WriteLine($"min = {numbers[0]}, max = {numbers[1]}");
    return;
}

int i = 0;
int j = numbers.Length - 1;

int min = numbers[i];
int max = numbers[j];
i++;
j--;

while (i <= j)
{
    if(numbers[i] > numbers[j])
    {
        if (numbers[j] < min) min = numbers[j];
        if (numbers[i] > max) max = numbers[i];
    }
    else
    {
        if (numbers[i] < min) min = numbers[i];
        if (numbers[j] > max) max = numbers[j];
    }
    i++;
    j--;
}

这是一个用 C# 编写的解决方案。我发现这种在两端燃烧蜡烛的方法是一个很好的解决方案。

于 2019-11-06T20:14:42.970 回答
1

成对比较最适合最少的比较

# Initialization # 
- if len(arr) is even, min = min(arr[0], arr[1]), max = max(arr[0], arr[1])
- if len(arr) is odd, min = min = arr[0], max = arr[0]

# Loop over pairs # 
- Compare bigger of the element with the max, and smaller with min,
- if smaller element less than min, update min, similarly with max.

比较总数 -

  • 对于 size = 奇数,3(n - 1) / 2 其中 n 是数组的大小
  • 对于大小 = 偶数,1 + 3*(n - 2)/2 = 3n/2 - 2

下面是上述伪代码的python代码

class Solution(object):
    def min_max(self, arr):
        size = len(arr)
        if size == 1:
            return arr[0], arr[0]

        if size == 2:
            return arr[0], arr[1]

        min_n = None
        max_n = None
        index = None
        if size % 2 == 0:       # One comparison
            min_n = min(arr[0], arr[1])
            max_n = max(arr[0], arr[1])
            st_index = 2

        else:
            min_n = arr[0]
            max_n = arr[0]
            st_index = 1
        for index in range(st_index, size, 2):
            if arr[index] < arr[index + 1]:
                min_n = min(arr[index], min_n)
                max_n = max(arr[index + 1], max_n)
            else:
                min_n = min(arr[index + 1], min_n)
                max_n = max(arr[index], max_n)

        return min_n, max_n
于 2020-10-17T02:03:57.407 回答
-5

只需循环一次数组,跟踪到目前为止的最大值和最小值。

于 2012-11-24T18:58:37.180 回答