在编码中查找缺少的整数
本文关键字:整数 查找 编码 | 更新日期: 2023-09-27 18:12:30
I need to "Find the minimal positive integer not occurring in a given sequence. "
A[0] = 1
A[1] = 3
A[2] = 6
A[3] = 4
A[4] = 1
A[5] = 2, the function should return 5.
Assume that:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
我用编码编写代码,但在许多情况下它不起作用,性能测试给出 0 %。请帮助我,我错在哪里。
class Solution {
public int solution(int[] A) {
if(A.Length ==0) return -1;
int value = A[0];
int min = A.Min();
int max = A.Max();
for (int j = min+1; j < max; j++)
{
if (!A.Contains(j))
{
value = j;
if(value > 0)
{
break;
}
}
}
if(value > 0)
{
return value;
}
else return 1;
}
}
编码给出除示例之外的所有错误,仅正值和负值。
编辑:添加了细节以更直接地回答您的实际问题。
"请帮帮我,我错在哪里。">
在正确性方面:考虑A = {7,2,5,6,3}
.给定 A
的内容,正确的输出是 1
,但是我们的算法将无法检测到这一点,因为 A.Min(( 会返回2
并且我们会从3
开始循环。在这种情况下,我们将返回4
;因为它是下一个缺失值。
像A = {14,15,13}
这样的事情也是如此。这里缺少的最小正整数再次1
,并且由于 13-15 的所有值都存在,因此 value
变量将保留其初始值 value=A[0]
,这将14
。
在性能方面:考虑A.Min()
、A.Max()
和A.Contains()
在幕后做什么;每一个都在完整地循环A
,在Contains
的情况下,我们对Min()
和我们能找到的最低正整数之间的每个值重复调用它。这将使我们远远超出Codility正在寻找的指定O(N)
性能。
相比之下,这是我能想到的最简单的版本,在Codility上应该得分为100%。请注意,我们只循环遍历A
一次,并且我们利用了一个允许我们使用ContainsKey
的Dictionary
;一种更快的方法,不需要遍历整个集合来查找值。
using System;
using System.Collections.Generic;
class Solution {
public int solution(int[] A) {
// the minimum possible answer is 1
int result = 1;
// let's keep track of what we find
Dictionary<int,bool> found = new Dictionary<int,bool>();
// loop through the given array
for(int i=0;i<A.Length;i++) {
// if we have a positive integer that we haven't found before
if(A[i] > 0 && !found.ContainsKey(A[i])) {
// record the fact that we found it
found.Add(A[i], true);
}
}
// crawl through what we found starting at 1
while(found.ContainsKey(result)) {
// look for the next number
result++;
}
// return the smallest positive number that we couldn't find.
return result;
}
}
获得满分的最简单解决方案是:
public int solution(int[] A)
{
int flag = 1;
A = A.OrderBy(x => x).ToArray();
for (int i = 0; i < A.Length; i++)
{
if (A[i] <= 0)
continue;
else if (A[i] == flag)
{
flag++;
}
}
return flag;
}
...1,000,000]。
public int solution(int[] array)
{
HashSet<int> found = new HashSet<int>();
for (int i = 0; i < array.Length; i++)
{
if (array[i] > 0)
{
found.Add(array[i]);
}
}
int result = 1;
while (found.Contains(result))
{
result++;
}
return result;
}
100% 的 C# 的微小版本
using System.Linq;
class Solution
{
public int solution(int[] A)
{
// write your code in C# 6.0 with .NET 4.5 (Mono)
var i = 0;
return A.Where(a => a > 0).Distinct().OrderBy(a => a).Any(a => a != (i = i + 1)) ? i : i + 1;
}
}
一个使用 C# 获得 100% 分数的简单解决方案
int Solution(int[] A)
{
var A2 = Enumerable.Range(1, A.Length + 1);
return A2.Except(A).First();
}
public class Solution {
public int solution( int[] A ) {
return Arrays.stream( A )
.filter( n -> n > 0 )
.sorted()
.reduce( 0, ( a, b ) -> ( ( b - a ) > 1 ) ? a : b ) + 1;
}
}
过滤掉负数似乎最容易。然后对流进行排序。然后减少它以得出答案。这是一种功能方法,但它的测试分数为 100/100。
使用此解决方案获得了 100% 的分数:https://app.codility.com/demo/results/trainingUFKJSB-T8P/
public int MissingInteger(int[] A)
{
A = A.Where(a => a > 0).Distinct().OrderBy(c => c).ToArray();
if (A.Length== 0)
{
return 1;
}
for (int i = 0; i < A.Length; i++)
{
//Console.WriteLine(i + "=>" + A[i]);
if (i + 1 != A[i])
{
return i + 1;
}
}
return A.Max() + 1;
}
JavaScript 解决方案,使用 Hash Table 且具有 O(n( 时间复杂度。
function solution(A) {
let hashTable = {}
for (let item of A) {
hashTable[item] = true
}
let answer = 1
while(true) {
if(!hashTable[answer]) {
return answer
}
answer++
}
}
C# 最简单的解决方案是:
int value = 1;
int min = A.Min();
int max = A.Max();
if (A.Length == 0) return value = 1;
if (min < 0 && max < 0) return value = 1;
List<int> range = Enumerable.Range(1, max).ToList();
List<int> current = A.ToList();
List<int> valid = range.Except(current).ToList();
if (valid.Count() == 0)
{
max++;
return value = max;
}
else
{
return value = valid.Min();
}
Considering that the array should start from 1 or if it needs to start from the minimum value than the Enumerable.range should start from Min
C 语言中的 Missing Integer 解
int solution(int A[], int N) {
int i=0,r[N];
memset(r,0,(sizeof(r)));
for(i=0;i<N;i++)
{
if(( A[i] > 0) && (A[i] <= N)) r[A[i]-1]=A[i];
}
for(i=0;i<N;i++)
{
if( r[i] != (i+1)) return (i+1);
}
return (N+1);
}
我的解决方案:
public static int solution()
{
var A = new[] { -1000000, 1000000 }; // You can try with different integers
A = A.OrderBy(i => i).ToArray(); // We sort the array first
if (A.Length == 1) // if there is only one item in the array
{
if (A[0]<0 || A[0] > 1)
return 1;
if (A[0] == 1)
return 2;
}
else // if there are more than one item in the array
{
for (var i = 0; i < A.Length - 1; i++)
{
if (A[i] >= 1000000) continue; // if it's bigger than 1M
if (A[i] < 0 || (A[i] + 1) >= (A[i + 1])) continue; //if it's smaller than 0, if the next integer is bigger or equal to next integer in the sequence continue searching.
if (1 < A[0]) return 1;
return A[i] + 1;
}
}
if (1 < A[0] || A[A.Length - 1] + 1 == 0 || A[A.Length - 1] + 1 > 1000000)
return 1;
return A[A.Length-1] +1;
}
class Solution {
public int solution(int[] A) {
int size=A.length;
int small,big,temp;
for (int i=0;i<size;i++){
for(int j=0;j<size;j++){
if(A[i]<A[j]){
temp=A[j];
A[j]=A[i];
A[i]=temp;
}
}
}
int z=1;
for(int i=0;i<size;i++){
if(z==A[i]){
z++;
}
//System.out.println(a[i]);
}
return z;
}
enter code here
}
在 C# 中,您可以通过使用内置库函数来解决此问题。对于非常大的整数,性能如何低
public int solution(int[] A)
{
var numbers = Enumerable.Range(1, Math.Abs(A.Max())+1).ToArray();
return numbers.Except(A).ToArray()[0];
}
如果您找到更好的解决方案,请告诉我性能明智
C# - MissingInteger
查找 1 - 1000.000 之间的最小缺失整数。
OP的假设发生
任务得分/正确性/性能:100%
using System;
using System.Linq;
namespace TestConsole
{
class Program
{
static void Main(string[] args)
{
var A = new int[] { -122, -5, 1, 2, 3, 4, 5, 6, 7 }; // 8
var B = new int[] { 1, 3, 6, 4, 1, 2 }; // 5
var C = new int[] { -1, -3 }; // 1
var D = new int[] { -3 }; // 1
var E = new int[] { 1 }; // 2
var F = new int[] { 1000000 }; // 1
var x = new int[][] { A, B, C, D, E, F };
x.ToList().ForEach((arr) =>
{
var s = new Solution();
Console.WriteLine(s.solution(arr));
});
Console.ReadLine();
}
}
// ANSWER/SOLUTION
class Solution
{
public int solution(int[] A)
{
// clean up array for negatives and duplicates, do sort
A = A.Where(entry => entry > 0).Distinct().OrderBy(it => it).ToArray();
int lowest = 1, aLength = A.Length, highestIndex = aLength - 1;
for (int i = 0; i < aLength; i++)
{
var currInt = A[i];
if (currInt > lowest) return lowest;
if (i == highestIndex) return ++lowest;
lowest++;
}
return 1;
}
}
}
获得 100% - C# 高效解决方案
public int solution (int [] A){
int len = A.Length;
HashSet<int> realSet = new HashSet<int>();
HashSet<int> perfectSet = new HashSet<int>();
int i = 0;
while ( i < len)
{
realSet.Add(A[i]); //convert array to set to get rid of duplicates, order int's
perfectSet.Add(i + 1); //create perfect set so can find missing int
i++;
}
perfectSet.Add(i + 1);
if (realSet.All(item => item < 0))
return 1;
int notContains =
perfectSet.Except(realSet).Where(item=>item!=0).FirstOrDefault();
return notContains;
}
class Solution {
public int solution(int[] a) {
int smallestPositive = 1;
while(a.Contains(smallestPositive)) {
smallestPositive++;
}
return smallestPositive;
}
}
嗯,现在这是一个新的赢家。至少在 C# 和我的笔记本电脑上。它比之前的冠军快 1.5-2 倍,比大多数其他解决方案快 3-10 倍。此解决方案的功能(或错误?(是它仅使用基本数据类型。还有100/100关于法典。
public int Solution(int[] A)
{
bool[] B = new bool[(A.Length + 1)];
for (int i = 0; i < A.Length; i++)
{
if ((A[i] > 0) && (A[i] <= A.Length))
B[A[i]] = true;
}
for (int i = 1; i < B.Length; i++)
{
if (!B[i])
return i;
}
return A.Length + 1;
}
简单的C++解决方案。无需额外的内存,时间执行顺序 O(N*log(N((:
int solution(vector<int> &A) {
sort (A.begin(), A.end());
int prev = 0; // the biggest integer greater than 0 found until now
for( auto it = std::begin(A); it != std::end(A); it++ ) {
if( *it > prev+1 ) break;// gap found
if( *it > 0 ) prev = *it; // ignore integers smaller than 1
}
return prev+1;
}
int[] A = {1, 3, 6, 4, 1, 2};
Set<Integer> integers = new TreeSet<>();
for (int i = 0; i < A.length; i++) {
if (A[i] > 0) {
integers.add(A[i]);
}
}
Integer[] arr = integers.toArray(new Integer[0]);
final int[] result = {Integer.MAX_VALUE};
final int[] prev = {0};
final int[] curr2 = {1};
integers.stream().forEach(integer -> {
if (prev[0] + curr2[0] == integer) {
prev[0] = integer;
} else {
result[0] = prev[0] + curr2[0];
}
});
if (Integer.MAX_VALUE == result[0]) result[0] = arr[arr.length-1] + 1;
System.out.println(result[0]);
惊讶,但这是一个很好的教训。林克很慢。我下面的答案让我得到 11%
public int solution (int [] A){
if (Array.FindAll(A, x => x >= 0).Length == 0) {
return 1;
} else {
var lowestValue = A.Where(x => Array.IndexOf(A, (x+1)) == -1).Min();
return lowestValue + 1;
}
}
我想我对此的看法有点不同,但得到了 100% 的评价。另外,我没有使用库:
public static int Solution(int[] A)
{
var arrPos = new int[1_000_001];
for (int i = 0; i < A.Length; i++)
{
if (A[i] >= 0)
arrPos[A[i]] = 1;
}
for (int i = 1; i < arrPos.Length; i++)
{
if (arrPos[i] == 0)
return i;
}
return 1;
}
public int solution(int[] A) {
// write your code in Java SE 8
Set<Integer> elements = new TreeSet<Integer>();
long lookFor = 1;
for (int i = 0; i < A.length; i++) {
elements.add(A[i]);
}
for (Integer integer : elements) {
if (integer == lookFor)
lookFor += 1;
}
return (int) lookFor;
}
我尝试在 C# 中使用递归而不是排序,因为我认为这样做会显示出更多的编码技能,但在扩展测试中,它在大型性能测试中表现不佳。假设最好只做简单的方法。
class Solution {
public int lowest=1;
public int solution(int[] A) {
// write your code in C# 6.0 with .NET 4.5 (Mono)
if (A.Length < 1)
return 1;
for (int i=0; i < A.Length; i++){
if (A[i]==lowest){
lowest++;
solution(A);
}
}
return lowest;
}
}
在javascript中的解决方案
function solution(A) {
// write your code in JavaScript (Node.js 8.9.4)
let result = 1;
let haveFound = {}
let len = A.length
for (let i=0;i<len;i++) {
haveFound[`${A[i]}`] = true
}
while(haveFound[`${result}`]) {
result++
}
return result
}
class Solution {
public int solution(int[] A) {
var sortedList = A.Where(x => x > 0).Distinct().OrderBy(x => x).ToArray();
var output = 1;
for (int i = 0; i < sortedList.Length; i++)
{
if (sortedList[i] != output)
{
return output;
}
output++;
}
return output;
}
}
你应该只使用HashSet,因为它的查找时间也是恒定的,而不是字典。代码更少更干净。
public int solution (int [] A){
int answer = 1;
var set = new HashSet<int>(A);
while (set.Contains(answer)){
answer++;
}
return answer;
}
此代码段应该可以正常工作。
using System;
using System.Collections.Generic;
public class Program
{
public static void Main()
{
int result = 1;
List<int> lst = new List<int>();
lst.Add(1);
lst.Add(2);
lst.Add(3);
lst.Add(18);
lst.Add(4);
lst.Add(1000);
lst.Add(-1);
lst.Add(-1000);
lst.Sort();
foreach(int curVal in lst)
{
if(curVal <=0)
result=1;
else if(!lst.Contains(curVal+1))
{
result = curVal + 1 ;
}
Console.WriteLine(result);
}
}
}