对于大小为N的数组,需要多少比较次数?
对于大小为N的数组,需要多少比较次数?
int[] int_array = {4, 6, 2, 9, 1, 7, 4, 2, 9, 0, 3, 6, 1, 6, 8};
int largst=int_array[0];
int second=int_array[0];
for (int i=0; i<int_array.length; i++){
if(int_array[i]>largst) {
second=largst;
largst=int_array[i];
}
else if(int_array[i]>second && int_array[i]<largst) {
second=int_array[i];
}
}
可以通过n + ceil(log n) - 2次比较完成。
解决方案: 需要n-1次比较才能得到最小值。
但是为了得到最小值,我们将构建一个锦标赛,其中每个元素将被分成一对,就像网球比赛一样,每轮的获胜者将前进。
这棵树的高度将是log n,因为我们每轮都会减半。
获取第二小的想法是它将被前面某一轮中的最小候选者击败。因此,我们需要在潜在的候选人中找到最小值(被最小值击败)。
潜在的候选人将是树的高度log n
因此,使用锦标赛树查找最小值的比较次数为n-1,查找第二小的比较次数为log n -1,总和为n + ceil(log n) - 2
以下是C++代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
using namespace std;
typedef pair<int,int> ii;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is for the case when x is 0 */
return x && (!(x&(x-1)));
}
// modified
int log_2(unsigned int n) {
int bits = 0;
if (!isPowerOfTwo(n))
bits++;
if (n > 32767) {
n >>= 16;
bits += 16;
}
if (n > 127) {
n >>= 8;
bits += 8;
}
if (n > 7) {
n >>= 4;
bits += 4;
}
if (n > 1) {
n >>= 2;
bits += 2;
}
if (n > 0) {
bits++;
}
return bits;
}
int second_minima(int a[], unsigned int n) {
// build a tree of size of log2n in the form of 2d array
// 1st row represents all elements which fights for min
// candidate pairwise. winner of each pair moves to 2nd
// row and so on
int log_2n = log_2(n);
long comparison_count = 0;
// pair of ints : first element stores value and second
// stores index of its first row
ii **p = new ii*[log_2n];
int i, j, k;
for (i = 0, j = n; i < log_2n; i++) {
p[i] = new ii[j];
j = j&1 ? j/2+1 : j/2;
}
for (i = 0; i < n; i++)
p[0][i] = make_pair(a[i], i);
// find minima using pair wise fighting
for (i = 1, j = n; i < log_2n; i++) {
// for each pair
for (k = 0; k+1 < j; k += 2) {
// find its winner
if (++comparison_count && p[i-1][k].first < p[i-1][k+1].first) {
p[i][k/2].first = p[i-1][k].first;
p[i][k/2].second = p[i-1][k].second;
}
else {
p[i][k/2].first = p[i-1][k+1].first;
p[i][k/2].second = p[i-1][k+1].second;
}
}
// if no. of elements in row is odd the last element
// directly moves to next round (row)
if (j&1) {
p[i][j/2].first = p[i-1][j-1].first;
p[i][j/2].second = p[i-1][j-1].second;
}
j = j&1 ? j/2+1 : j/2;
}
int minima, second_minima;
int index;
minima = p[log_2n-1][0].first;
// initialize second minima by its final (last 2nd row)
// potential candidate with which its final took place
second_minima = minima == p[log_2n-2][0].first ? p[log_2n-2][1].first : p[log_2n-2][0].first;
// minima original index
index = p[log_2n-1][0].second;
for (i = 0, j = n; i <= log_2n - 3; i++) {
// if its last candidate in any round then there is
// no potential candidate
if (j&1 && index == j-1) {
index /= 2;
j = j/2+1;
continue;
}
// if minima index is odd, then it fighted with its index - 1
// else its index + 1
// this is a potential candidate for second minima, so check it
if (index&1) {
if (++comparison_count && second_minima > p[i][index-1].first)
second_minima = p[i][index-1].first;
}
else {
if (++comparison_count && second_minima > p[i][index+1].first)
second_minima = p[i][index+1].first;
}
index/=2;
j = j&1 ? j/2+1 : j/2;
}
printf("-------------------------------------------------------------------------------\n");
printf("Minimum : %d\n", minima);
printf("Second Minimum : %d\n", second_minima);
printf("comparison count : %ld\n", comparison_count);
printf("Least No. Of Comparisons (");
printf("n+ceil(log2_n)-2) : %d\n", (int)(n+ceil(log(n)/log(2))-2));
return 0;
}
int main()
{
unsigned int n;
scanf("%u", &n);
int a[n];
int i;
for (i = 0; i < n; i++)
scanf("%d", &a[i]);
second_minima(a,n);
return 0;
}
假设空间不重要,这是我能够缩小的最小值。在最坏情况下需要2*n次比较,在最好情况下需要n次比较:
arr = [ 0, 12, 13, 4, 5, 32, 8 ]
max = [ -1, -1 ]
for i in range(len(arr)):
if( arr[i] > max[0] ):
max.insert(0,arr[i])
elif( arr[i] > max[1] ):
max.insert(1,arr[i])
print max[1]
这是sdcvvc在C++11中提供的可接受解决方案。
#include <algorithm>
#include <iostream>
#include <vector>
#include <cassert>
#include <climits>
using std::vector;
using std::cout;
using std::endl;
using std::random_shuffle;
using std::min;
using std::max;
vector<int> create_tournament(const vector<int>& input) {
// make sure we have at least two elements, so the problem is interesting
if (input.size() <= 1) {
return input;
}
vector<int> result(2 * input.size() - 1, -1);
int i = 0;
for (const auto& el : input) {
result[input.size() - 1 + i] = el;
++i;
}
for (uint j = input.size() / 2; j > 0; j >>= 1) {
for (uint k = 0; k < 2 * j; k += 2) {
result[j - 1 + k / 2] = min(result[2 * j - 1 + k], result[2 * j + k]);
}
}
return result;
}
int second_smaller(const vector<int>& tournament) {
const auto& minimum = tournament[0];
int second = INT_MAX;
for (uint j = 0; j < tournament.size() / 2; ) {
if (tournament[2 * j + 1] == minimum) {
second = min(second, tournament[2 * j + 2]);
j = 2 * j + 1;
}
else {
second = min(second, tournament[2 * j + 1]);
j = 2 * j + 2;
}
}
return second;
}
void print_vector(const vector<int>& v) {
for (const auto& el : v) {
cout << el << " ";
}
cout << endl;
}
int main() {
vector<int> a;
for (int i = 1; i <= 2048; ++i)
a.push_back(i);
for (int i = 0; i < 1000; i++) {
random_shuffle(a.begin(), a.end());
const auto& v = create_tournament(a);
assert (second_smaller(v) == 2);
}
return 0;
}
试一下这个。
max1 = a[0].
max2.
for i = 0, until length:
if a[i] > max:
max2 = max1.
max1 = a[i].
#end IF
#end FOR
return min2.
它应该像魔法一样运行。复杂度低。
这里是Java代码。
int secondlLargestValue(int[] secondMax){
int max1 = secondMax[0]; // assign the first element of the array, no matter what, sorted or not.
int max2 = 0; // anything really work, but zero is just fundamental.
for(int n = 0; n < secondMax.length; n++){ // start at zero, end when larger than length, grow by 1.
if(secondMax[n] > max1){ // nth element of the array is larger than max1, if so.
max2 = max1; // largest in now second largest,
max1 = secondMax[n]; // and this nth element is now max.
}//end IF
}//end FOR
return max2;
}//end secondLargestValue()
#include<stdio.h>
main()
{
int a[5] = {55,11,66,77,72};
int max,min,i;
int smax,smin;
max = min = a[0];
smax = smin = a[0];
for(i=0;i<=4;i++)
{
if(a[i]>max)
{
smax = max;
max = a[i];
}
if(max>a[i]&&smax<a[i])
{
smax = a[i];
}
}
printf("the first max element z %d\n",max);
printf("the second max element z %d\n",smax);
}
使用计数排序,然后从索引0开始向末尾查找第二大的元素。至少应该有1次比较,最多n-1
次(当只有一个元素时!)。
function findSecondLargeNumber(arr){
var fLargeNum = 0;
var sLargeNum = 0;
for(var i=0; i<arr.length; i++){
if(fLargeNum < arr[i]){
sLargeNum = fLargeNum;
fLargeNum = arr[i];
}else if(sLargeNum < arr[i]){
sLargeNum = arr[i];
}
}
return sLargeNum;
}
var myArray = [799, -85, 8, -1, 6, 4, 3, -2, -15, 0, 207, 75, 785, 122, 17];
参考:http://www.ajaybadgujar.com/finding-second-largest-number-from-array-in-javascript/
我已经阅读了以上所有帖子,但我相信锦标赛算法的实现是最好的方法。让我们考虑@Gumbo发布的以下算法
largest := numbers[0];
secondLargest := null
for i=1 to numbers.length-1 do
number := numbers[i];
if number > largest then
secondLargest := largest;
largest := number;
else
if number > secondLargest then
secondLargest := number;
end;
end;
end;
如果我们要在数组中找到第二大的数字,那么这个算法非常好用。它需要(2n-1)次比较。但是如果你想要计算第三大的数字或者第k大的数字呢?上面的算法就不行了,你需要使用另一种方法。
所以,我认为锦标赛算法是最好的方法,这里是链接。