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，我可以循环数组两次并在最坏的情况下使用~2n比较，但我想做得更好。
答案 0 :(得分：65)
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)
通过这种方式，您可以对2个元素进行3次比较，相当于3N/2元素的N总比较。
答案 1 :(得分：15)
尝试通过srbh.kmr改进答案。假设我们有序列：
A = [a1, a2, a3, a4, a5]
比较a1＆amp; a2并计算min12，max12：
if (a1 > a2)
min12 = a2
max12 = a1
else
min12 = a1
max12 = a2
同样计算min34，max34。由于a5是唯一的，因此请保持原样......
现在比较min12＆amp; min34并计算min14，类似地计算max14。最后比较min14＆amp; a5计算min15。同样地计算max15。
总共只有6次比较！
此解决方案可以扩展为任意长度的数组。可能可以通过类似的合并排序方法来实现（将数组分成两半并为每一半计算min max。
UPDATE：以下是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）来确定比较的确切数量。但是很难看出人们如何能够在这么多比较之下。
更新：我们可以计算出如下比较次数：
在这个计算树的底部，我们从原始数组中形成整数对。所以我们有N / 2个叶子节点。对于每个叶节点，我们只进行了1次比较。
通过引用perfect-binary-tree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，我们进行了N - 2次比较。与叶节点处的N / 2比较一起，我们进行了(3N / 2) - 2次总比较。
所以，这可能是他的回答中隐含的解决方案srbh.kmr。
答案 2 :(得分：5)
去分而治之！
1,3,2,5
对于这个发现min，max将进行6次比较
但除以它们
1,3 ---＆gt;将在一次比较中给出最小1和最大3 2,5 ---＆gt;将在一次比较中给出min 2和max 5
现在我们可以比较两分钟（1,2） - ＆gt;将最终分钟作为1（一个比较） 同样两个最大（3,5）---＆gt;将最终的最大值设为5（一个比较）
所以完全进行了四次比较
答案 3 :(得分：4)
一种稍微不同的方法，它使用整数运算而不是比较（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;
}
答案 4 :(得分：3)
蛮力更快！
我希望有人向我展示我的方式的错误，但是，......，
我将蛮力方法的实际运行时间与（更美丽的）递归分而治之比较。典型结果（每个函数10,000,000次调用）：
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个项目的阵列，蛮力方法的速度提高了2.9倍，而1,000,000个项目的数组则快了3.4倍。
显然，比较次数不是问题，但可能是重新分配的次数，以及调用递归函数的开销（这可能解释为什么1,000,000个值的运行速度慢于10个值）。
警告：我在VBA而不是C中这样做，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
'2014.07.02
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
答案 5 :(得分：3)
在阅读问题和答案后，我决定尝试一些版本（在C＃中） 我认为最快的将是Anton Knyazyev的一个（免费分支）， 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＆＃34;分支预测器＆＃34;做得很好，
每次迭代机会，找到新的最小值（或最大值），减少，
所以预测会变得更好
总而言之，它只是一个简单的测试。我确实知道我的结论可能是：
-premature。
- 不适用于不同的平台
让他们说他们是指示性的
编辑：收支平衡点minmax0，minmax3：~100项，
10,000项：minmax3~minmax0的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;
}
}
答案 6 :(得分：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)).
答案 7 :(得分：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);
}
}
答案 8 :(得分：1)
我的分歧＆amp;到目前为止用java征服方法：
service.getRates()
.map(ratesToMapWithEmptyValues)
答案 9 :(得分：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)));
}
答案 10 :(得分：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）复杂性中完成!!!
答案 11 :(得分：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。
答案 12 :(得分：1)
成对比较最适合进行最小比较
// The following block might slightly improve the execution time;
// Can be removed;
static const auto __optimize__ = []() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
return 0;
}();
// Most of headers are already included;
// Can be removed;
#include <cstdint>
#include <string>
#include <sstream>
static const struct Solution {
static const int isPrefixOfWord(
const std::string sentence,
const std::string_view search_word
) {
std::basic_stringstream stream_sentence(sentence);
std::size_t index = 1;
std::string word;
while (stream_sentence >> word) {
if (!word.find(search_word)) {
return index;
}
++index;
}
return -1;
}
};
比较总数-
以下是上述伪代码的python代码
# 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.
答案 13 :(得分：-3)
只需在数组上循环一次，跟踪到目前为止的最大值和最小值。