Quick queue application
Ideas
-
The K-th largest number is required, so use partitions sorted from largest to smallest
-
In findK, first partition produces a result, and then compares the relative size of K and the position (pos-left+1) to continue left divide and conquer/right divide and conquer
-
For the case of K>M in the findK function, left needs to be changed, and the value of K needs to be adjusted for the current recursive position of left
- Description
- M is the divide and conquer point, which is pos+1-left (because left and pos+1 are relative to 0)
- K is also relative to left, so it needs to be updated continuously
0 left M(=pos+1-left) K To 0 position 0 left M+left/pos+1 Position left 0 M - M (relative to 0 is pos+1) should be used as the new left point
- 则newK=K-M=K-pos-1+left
- FindK(arr, pos+1 , right , KM ) is the distance relative to 0 in bold, and the distance relative to the current left in italics
- Description
Code
#include <stdio.h>
int partition(int arr[],int left,int right){
//注意是第K大,所以从大到小
int temp=arr[left];
while(left<right){
while(arr[right]<=temp&&left<right) right--;
arr[left]=arr[right];//注意要在这里就解决,在最后right会改变
while(arr[left]>temp&&left<right) left++;
arr[right]=arr[left];
}
arr[left]=temp;
//在此处左侧的都比temp小,右侧的都比temp大
return left;//需要返回下标作为下一次quicksort的位置,类似mid
}
int findK(int arr[],int left,int right,int K){
if(left<=right){
int pos=partition(arr,left,right);
int M=pos-left+1;//从left起第M大的数
if(K==M)
return arr[pos];
else if(K<M)
return findK(arr,left,pos-1,K);
else
return findK(arr,pos+1,right,K-M);//注意转化为K-M=K-pos+left-1(相对pos+1来说就是K-left,仍为最初始第K大数)
}
return 0;
}
int main(){
int n=0,K=0;
scanf("%d%d",&n,&K);
int arr[1000005]={
0};
for(int i=0;i<n;i++)
scanf("%d",&arr[i]);
int res=findK(arr,0,n-1,K);
printf("%d\n",res);
}