3dmax寻找丢失贴图

Problem statement

问题陈述

We are given a list of n-1 integers and the integers range from 1 to n. There are no duplicates in the list. Only one of the integer from 1 to n is missing. We need to find the missing number in O(n) time complexity.

我们给出了n-1个整数的列表，这些整数的范围是1到n。 列表中没有重复项。 从1到n的整数中只有一个丢失。 我们需要找到O(n)时间复杂度中的缺失数 。

Solution

解

The simplest solution is to take each number from 1 to n and to check whether it exists in the array or not. But such approach has O(n²) time complexity. Thus we need to find some advance algorithm to reduce the time complexity.

最简单的解决方案是将每个数字从1取到n，并检查它是否存在于数组中。 但是这种方法具有O(n ² )时间复杂度 。 因此，我们需要找到一些先进的算法来减少时间复杂度。

The algorithm is illustrated below:

该算法如下图所示：

1. XOR all the elements presented in the array. Let the result be x.

   对数组中显示的所有元素进行XOR 。 令结果为x 。

2. XOR all numbers from 1 to n. Let the result be y.

   将所有数字从1到n进行 XOR 。 令结果为y 。

3. XORing x & y gives the missing number.

   对 x和y 进行XORing运算得出缺失的数字。

C code for implementation

实现的C代码

#include<stdio.h>
#include<stdlib.h>
int missingNo(int* a, int n){int x=0,y=0;
for(int i=0;i<n-1;i++){//xoring all elements
x^=a[i];
}
for(int i=1;i<=n;i++){//xoring 1 to n
y^=i;
}
//xoring x & y outputs missing number
return x^y;
}
int main()
{int n,x,count=0;
printf("enter your range\n");
scanf("%d",&n);
printf("enter elements leaving one nummber in the range 1 to n\n");
// dynamic array created for n-1 elements
int* a=(int*)(malloc(sizeof(int)*(n-1)));
for(int i=0;i<n-1;i++){scanf("%d",&a[i]);
}
// function to check duplicate exists or not
printf("\nthe missing number is %d\n",missingNo(a,n));
return 0;
}

Output

输出量

enter your range
5
enter elements leaving one nummber in the range 1 to n
2
4
3
5
the missing number is 1

Time complexity: O(n) (for scanning the array)

时间复杂度： O(n)(用于扫描阵列)

Space complexity: O(1)

空间复杂度： O(1)

翻译自: https://www.includehelp.com/algorithms/finding-the-missing-number.aspx

3dmax寻找丢失贴图