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法律研究查找序列中第 k 大元素的 C++ 程序
在这个程序中，我们需要提取序列中的第 K 个最大元素。采用最大堆法解决此问题可以提升此技术的的时间复杂度。此程序的时间复杂度为 O(n + k*log(n))。
算法
Begin Send the max of the heap to the end of the sequence. Heapify the remaining sequence. Repeat the process for ‘k’ times. Print the final state of the array. Print the max from the heap extracted from kth iteration as a result. End.
示例代码
#include <iostream>
using namespace std;
void MaxHeapify(int a[], int i, int n) {
int j, t;
t = a[i];
j = 2*i;
while (j <= n) {
if (j < n && a[j+1] > a[j])
j = j+1;
if (t > a[j])
break;
else if (t <= a[j]) {
a[j/2] = a[j];
j = 2*j;
}
}
a[j/2] = t;
return;
}
void Build_MaxHeapify(int a[], int n) {
int i;
for(i = n/2; i >= 1; i--)
MaxHeapify(a, i, n);
}
int main() {
int n, i, temp, k;
cout<<"\nEnter the number of data element to be sorted: ";
cin>>n;
n++;
int arr[n];
for(i = 1; i < n; i++) {
cout<<"Enter element "<<i<<": ";
cin>>arr[i];
}
cout<<"\nEnter the k value: ";
cin>>k;
Build_MaxHeapify(arr, n-1);
for(i = n-1; i >= n-k; i--) {
temp = arr[i];
arr[i] = arr[1];
arr[1] = temp;
MaxHeapify(arr, 1, i - 1);
}
cout<<"\nAfter max-heapify the given array "<<k<<" times the array state is: ";
for(i = 1; i < n; i++)
cout<<"->"<<arr[i];
cout<<"\n\nThe "<<k<<"th largest element is: "<<arr[n-k];
return 0;
}输出
Enter the number of data element to be sorted: 5 Enter element 1: 20 Enter element 2: 10 Enter element 3: 30 Enter element 4: 70 Enter element 5: 60 Enter the k value: 2 After max-heapify the given array 2 times the array state is: ->30->20->10->60->70 The 2th largest element is: 60
更新于： 30-07-2019
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