Arrays 给定一个整数数组,在线性时间和常量空间中查找第一个缺失的正整数
换句话说,找到数组中不存在的最小正整数。数组也可以包含重复数和负数。 这个问题是Stripe在节目采访中提出的。我设计了一个解决方案,如下所示:Arrays 给定一个整数数组,在线性时间和常量空间中查找第一个缺失的正整数,arrays,algorithm,sorting,array-algorithms,Arrays,Algorithm,Sorting,Array Algorithms,换句话说,找到数组中不存在的最小正整数。数组也可以包含重复数和负数。 这个问题是Stripe在节目采访中提出的。我设计了一个解决方案,如下所示: #include<bits/stdc++.h> using namespace std; int main(){ int arr[]={1,-1,-5,-3,3,4,2,8}; int size= sizeof(arr)/sizeof(arr[0]); sort(arr, arr+size); int mi
#include<bits/stdc++.h>
using namespace std;
int main(){
int arr[]={1,-1,-5,-3,3,4,2,8};
int size= sizeof(arr)/sizeof(arr[0]);
sort(arr, arr+size);
int min=1;
for(int i=0; i<size; i++){
if(arr[i]>min) break;
if(arr[i]==min) min=min+1;
}
cout<<min;
return 0;
}
#包括
使用名称空间std;
int main(){
int arr[]={1,-1,-5,-3,3,4,2,8};
int size=sizeof(arr)/sizeof(arr[0]);
排序(arr,arr+大小);
int min=1;
对于(int i=0;imin)中断;
如果(arr[i]==min)min=min+1;
}
cout假设数组可以修改
我们将数组分成两部分,第一部分仅由正数组成。假设起始索引为0
,结束索引为end
(独占)
我们从索引0
到end
遍历数组。我们取该索引处元素的绝对值-假设值为x
如果x>end
我们什么也不做
如果不是,我们将索引x-1
处元素的符号设为负值。(澄清:我们不切换符号。如果值为正,它将变为负。如果值为负,它将保持为负。在伪代码中,这类似于If(arr[x-1]>0)arr[x-1]=-arr[x-1]
而不是arr[x-1]=-arr[x-1]
)
最后,我们再次从索引0
到end
遍历数组。如果在某个索引处遇到正元素,我们将输出index+1
。这就是答案。但是,如果我们没有遇到任何正元素,则表示数组中出现整数1
到end
。我们输出end+1
也可能是所有数字都是非正数,使得end=0
。输出end+1=1
保持正确
所有步骤都可以在O(n)
时间和O(1)
空间中完成
示例:
Initial Array: 1 -1 -5 -3 3 4 2 8
Step 1 partition: 1 8 2 4 3 | -3 -5 -1, end = 5
在步骤2中,我们更改正数的符号,以跟踪哪些整数已经出现。例如,这里array[2]=-2<0
,它表明2+1=3
已经在数组中出现。基本上,如果i+1
在数组中,我们将索引为i
的元素的值更改为负值
Step 2 Array changes to: -1 -8 -2 -4 3 | -3 -5 -1
在步骤3中,如果某个值array[index]
为正,则表示在步骤2中没有找到任何值index+1
的整数
Step 3: Traversing from index 0 to end, we find array[4] = 3 > 0
The answer is 4 + 1 = 5
PMCarpan的算法有效
我认为您的方法是可行的,但是您应该指定您正在执行的排序类型,以便很清楚它是线性排序,而不一定是整个数组的完整排序。这将导致O(N)时间,而不使用任何空间
在扫描数组时,如果当前索引中的值小于数组的长度,请扫描数组,然后将其与该索引中当前的值交换。您必须继续交换,直到在每个索引处交换不再有意义为止。然后在最后再进行一次扫描,直到找到不正确的索引
这里有一些可以工作的python代码,尽管python不是做这类事情的地方,哈哈
def sortOfSort(arr) :
for index in range(len(arr)) :
checkValue = arr[index]
while(checkValue > 0 and checkValue != index and checkValue < len(arr) and arr[checkValue] != checkValue) :
arr[index] = arr[checkValue]
arr[checkValue] = checkValue
checkValue = arr[index]
return arr[1:] + [arr[0]]
def findFirstMissingNumber(arr) :
for x in range(len(arr)) :
if (x+1 != arr[x]) :
return x+1
return len(arr) + 1
def排序(arr):
对于范围内的索引(len(arr)):
checkValue=arr[索引]
而(checkValue>0和checkValue!=索引和checkValue
返回arr[1:]部分是因为根据您的描述,我们没有将零作为起点。这里是一个C实现
输入
/*
如何工作:
[if(abs(arr[i])-10)
arr[abs(arr[i])-1]=-arr[abs(arr[i])-1];]
before:arr={7,3,4,5,5,3,2}
i==0:arr[0]=7
arr[7-1]为2>0~>否定
arr={7,3,4,5,5,3,-2}
i==1:arr[1]=3
arr[3-1]为4>0~>否定
arr={7,3,-4,5,5,3,-2}
i==2:arr[2]是用于索引的-4~>abs
arr[4-1]为5>0~>否定
arr={7,3,-4,-5,5,3,-2}
i==3:arr[3]是用于索引的-5~>abs
arr[5-1]为5>0~>否定
arr={7,3,-4,-5,-5,3,-2}
i==4:arr[4]是用于索引的-5~>abs
arr[5-1]为-5<0~>将abs(-5)打印为副本
i==5:arr[5]是3
arr[3-1]为-4<0~>将abs(3)打印为副本
i==6:arr[6]是用于索引的-2~>abs
arr[2-1]为3>0~>否定
arr={7,-3,-4,-5,-5,3,-2}
正项索引:0、5~>1和6不在原始数组中
负项索引:原始数组中的1,2,3,4,6~>2,3,4,5,7
*/
#返回包含正数的片段
def FindPositiveSubar(arr):
负指数=0
如果i在范围内(len(arr)):
如果arr[i]0:
正耳环[索引]*=-1
对于范围(l)中的i:
如果正耳环[i]>0:
返回i+1
返回l+1
如果名称=“\uuuuu main\uuuuuuuu”:
arr=[input().strip().split()中x的int(x)]
positiveSubArr=FindPositiveSubArr(arr)
打印(findmisingpositive(positiveSubArr))
我使用python3中的set解决了这个问题。它非常简单。
时间复杂度:O(n)
记住:成员签入集为O(1)
我并没有详细测试它,但对于排序数组,这里是我将如何处理它的,欢迎任何改进。
约束:
Initial Array: 1 -1 -5 -3 3 4 2 8
Step 1 partition: 1 8 2 4 3 | -3 -5 -1, end = 5
- 线性时间
- 恒定空间
solution:
start with lowest positive integer (i.e. lpi <- 1)
while parsing the array, if lpi is already in the array, increment it
如果数组i
/*
How work :
[if(abs(arr[i]) - 1 < size && arr[ abs(arr[i]) - 1] > 0)
arr[ abs(arr[i]) - 1] = -arr[ abs(arr[i]) - 1];]
before: arr = { 7, 3, 4, 5, 5, 3, 2}
i == 0: arr[0] = 7
arr[7-1] is 2 > 0 ~> negate
arr = { 7, 3, 4, 5, 5, 3, -2}
i == 1: arr[1] = 3
arr[3-1] is 4 > 0 ~> negate
arr = { 7, 3, -4, 5, 5, 3, -2}
i == 2: arr[2] is -4 ~> abs for indexing
arr[4-1] is 5 > 0 ~> negate
arr = { 7, 3, -4,-5, 5, 3, -2}
i == 3: arr[3] is -5 ~> abs for indexing
arr[5-1] is 5 > 0 ~> negate
arr = { 7, 3, -4, -5, -5, 3, -2}
i == 4: arr[4] is -5 ~> abs for indexing
arr[5-1] is -5 < 0 ~> print abs(-5) as duplicate
i == 5: arr[5] is 3
arr[3-1] is -4 < 0 ~> print abs(3) as duplicate
i == 6: arr[6] is -2 ~> abs for indexing
arr[2-1] is 3 > 0 ~> negate
arr = { 7, -3, -4, -5, -5, 3, -2}
indices of positive entries: 0, 5 ~> 1 and 6 not in original array
indices of negative entries: 1, 2, 3, 4, 6 ~> 2, 3, 4, 5, 7 in original array
*/
#Returns a slice containing positive numbers
def findPositiveSubArr(arr):
negativeIndex = 0
if i in range(len(arr)):
if arr[i] <=0:
arr.insert(negativeIndex, arr.pop(i))
negativeIndex += 1
return arr[negativeIndex:]
#Returns the first missing positive number
def findMissingPositive(positiveArr):
l = len(positiveArr)
for num in positiveArr:
index = abs(num) - 1
if index < 1 and positiveArr[index] > 0:
positiveArr[index] *= -1
for i in range(l):
if positiveArr[i] > 0:
return i+1
return l+1
if __name__ == "__main__":
arr = [int(x) for x in input().strip().split()]
positiveSubArr = findPositveSubArr(arr)
print(findMissingPositive(positiveSubArr))
def first_missing_positive_integer(arr):
arr = set(arr)
for i in range(1, len(arr)+2):
if i not in arr:
return i
solution:
start with lowest positive integer (i.e. lpi <- 1)
while parsing the array, if lpi is already in the array, increment it
def find_lpi(arr):
lpi = 1
for i in arr:
if lpi == i:
lpi += 1
return lpi
def find_lpi(arr):
x = [0 for x in range(max(arr)+1)]
for i in arr:
x[i] = 1
for i in range(1,len(x)):
if x[i] ==0:
return i
return len(x)
public static int Missing(int[] a)
{
// the idea is to put all values in array on their ordered place if possible
for (int i = 0; i < a.Length; i++)
{
CheckArrayAtPosition(a, i);
}
for (int i = 0; i < a.Length; i++)
if (a[i] != i + 1)
return i + 1;
return a.Length + 1;
}
private static void CheckArrayAtPosition(int[] a, int i)
{
var currentValue = a[i];
if (currentValue < 1) return; // do not touch negative values because array indexes are non-negative
if (currentValue > a.Length) return; // do not touch values that are bigger than array length because we will not locate them anyway
if (a[currentValue - 1] == currentValue) return; // do not need to change anything because index contain correct value already
Swap(a, i, currentValue - 1);
CheckArrayAtPosition(a, i); // now current position value is updated so we need to check current position again
}
private static void Swap(int[] a, int i, int j)
{
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
public int FindMissing(){
var list = new int[] { 6, -6, 4, 5 };
list = list.OrderBy(x => x).ToArray();
var maxValue = 0;
for (int i = 0; i < list.Length; i++)
{
if (list[i] <= 0)
{
continue;
}
if (i == list.Length - 1 ||
list[i] + 1 != list[i + 1])
{
maxValue = list[i] + 1;
break;
}
}
return maxValue;
}
def missing_int(nums: MutableSequence[int]) -> int:
# If empty array or doesn't have 1, return 1
if not next((x for x in nums if x == 1), 0):
return 1
lo: int = 0
hi: int = len(nums) - 1
i: int = 0
pivot: int = 1
while i <= hi:
if nums[i] < pivot:
swap(nums, i, hi)
hi -= 1
elif nums[i] > pivot:
swap(nums, i, lo)
i += 1
lo += 1
else:
i += 1
x = 0
while x <= hi: # hi is the index of the last positive number
y: int = abs(nums[x])
if 0 < y <= hi + 1 and nums[y - 1] > 0: # Don't flip sign if already negative
nums[y - 1] *= -1
x += 1
return next((i for i, v in enumerate(nums[:hi + 1]) if v >= 0), x) + 1
def test_missing_int(self):
assert func.missing_int([1, 2, 1, 0]) == 3
assert func.missing_int([3, 4, -1, 1]) == 2
assert func.missing_int([7, 8, 9, 11, 12]) == 1
assert func.missing_int([1]) == 2
assert func.missing_int([]) == 1
assert func.missing_int([0]) == 1
assert func.missing_int([2, 1]) == 3
assert func.missing_int([-1, -2, -3]) == 1
assert func.missing_int([1, 1]) == 2
assert func.missing_int([1000, -1]) == 1
assert func.missing_int([-10, -3, -100, -1000, -239, 1]) == 2
assert func.missing_int([1, 1]) == 2
def segregate(arr):
length = len(arr)
neg_index = length
for i, value in enumerate(arr):
if(value < 1 and neg_index == length):
neg_index = i
if(neg_index != length and value >= 1):
temp = arr[i]
arr[i] = arr[neg_index]
arr[neg_index] = temp
neg_index += 1
return arr[:neg_index]
def missingPositiveNumber(arr):
arr = segregate(arr)
length = len(arr)
for i, value in enumerate(arr):
if(value - 1 < l):
arr[abs(value) - 1] = -(abs(arr[abs(value) - 1]))
for i, value in enumerate(arr):
if(value > 0):
return i + 1
return length + 1
print(missingPositiveNumber([1, -1, 2, 3]))
private static int minimum_positive_integer(int[] arr) {
int i = 0;
int j = arr.length - 1;
//splitting array
while (i < j) {
if (arr[i] > 0) {
i++;
}
if (arr[j] <= 0) {
j--;
}
if (arr[i] <= 0 && arr[j] > 0) {
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
i++;
j--;
}
}
int len_positive = i;
if (arr[i] > 0) len_positive++;
for (i = 0; i < len_positive; i++) {
int abs = Math.abs(arr[i]);
if (abs <= len_positive) {
int index = abs - 1;
arr[index] = -abs;
}
}
for (i = 0; i < len_positive; i++) {
if(arr[i] > 0) return i + 1;
}
return len_positive + 1;
}
def lowest_positive(lista):
result = 0
dict = {}
for i in lista:
if i <= 0:
continue
if i in dict:
continue
else:
dict[i] = i
if result == 0:
result = result +1
if result < i:
continue
result = result +1
while result in dict:
result = result +1
return result
lista = [5, 3, 4, -1, 1, 2]
lista = [1,2,3,4,5]
lista = [3, 4, -1, 1]
lista = [2, 3, 4, 1]
lista = [1,0]
lowest_positive(lista)
def missing_positive_integer(my_list):
max_value = max(my_list)
my_list = [num for num in range(1,max(my_list)) if num not in my_list]
if len(my_list) == 0:
my_list.append(max_value+1)
return min(my_list)
my_list = [1,2,3,4,5,8,-1,-12,-3,-4,-8]
missing_positive_integer(my_list)