查找数组中的前 n 个最大元素 [英] Finding the first n largest elements in an array

问题描述

我有一个包含唯一元素的数组.我需要以尽可能低的复杂度找出数组中的前 n 个最大元素.到目前为止我能想到的解决方案的复杂度为 O(n^2).

I have got an array containing unique elements. I need to find out the first n largest elements in the array in the least complexity possible. The solution that I could think of so far has a complexity of O(n^2).

    int A[]={1,2,3,8,7,5,3,4,6};
    int max=0;
    int i,j;
    int B[4]={0,0,0,0,};//where n=4;
     for(i=0;i<A.length();i++)
       {
         if(A[i]>max)
          max=A[i];
       }
     B[0]=max;
     for(i=1;i<n;i++){
       max=0;
       for(j=0;j<A.length();j++){
         if(A[j]>max&&A[j]<B[i-1])
            max=A[j];
       }
        B[i]=max;
     }

请，如果有人能想出一个更简单的解决方案，我将不胜感激.而且我不打算改变原来的数组！！

Please, if anyone can come up with a better solution which involves less complexity, I will be highly grateful. And I don't intend to change the original array!!

推荐答案

找到第 k 个最大的元素，使用 selection算法.
接下来，迭代数组并找到所有大于/等于它的元素.

Find the kth biggest element, using selection algorithm.
Next, iterate the array and find all elements which are larger/equal it.

复杂度: O(n) 用于选择，O(n) 用于迭代，所以总和也是 O(n)

complexity: O(n) for selection and O(n) for iterating, so the total is also O(n)

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