# Reading diary-Linear time Sorting algorithm

Source: Internet
Author: User

In the worst case, any comparison sorting algorithm requires an O (NLGN) comparison.

However, under the specified conditions, the sequencing algorithm of linear time can make the sorting complete in O (n) time.

Count sort

Suppose that each of the N input elements is an integer within the range of 0 to K, where K is an integer. K=o (n), the sort run time is O (n).

Main ideas:

Create an array of length k, and use the value of the corresponding input array A as an index to count the occurrences of each subscript in the K array.

The code is as follows:

voidCounting_sort (int* Intarr,int* Outarr,intKintLen) {    int* CARR =New int[K]; memset (CARR,0, k*sizeof(int)); //Store the number of occurrences of each value with Carr     for(intI=0; i<len; ++i) {Carr[intarr[i]]+=1; }        //stores the current value exists, the size of the number of bits     for(intI=1; i<k; ++i) {Carr[i]=carr[i]+carr[i-1]; }         for(inti=len-1; i>=0; --i) {outarr[carr[intarr[i] ]=Intarr[i]; Carr[intarr[i]]-=1; }        Delete[] cArr;}

Base sort

Suppose you sort n d-bit integers, from low to high, D-rounds. Because each bit value is 0~9, it is necessary to sort the ordering method of the D-Times K to 9.

Thus the total time of the Cardinal sort is O (d (n+k)).

The code is as follows:

typedefstruct_basevar{intidx; int var;} Basevar;//Adjusted count sortvoidCOUNTING_SORT_EX (basevar* intarr,basevar* Outarr,intKintLen) {    int* CARR =New int[K]; memset (CARR,0, k*sizeof(int)); //Store the number of occurrences of each value with Carr     for(intI=0; i<len; ++i) {carr[intarr[i].var] +=1; }        //stores the current value exists, the size of the number of bits     for(intI=1; i<k; ++i) {Carr[i]=carr[i]+carr[i-1]; }         for(inti=len-1; i>=0; --i) {outarr[carr[intarr[i].var]].var=intarr[i].var; Carr[intarr[i]]-=1; }        Delete[] cArr;}//voidRadix_sort (int* Intarr,intDintLen) {Basevar* Vararr =NewBasevar[len]; Basevar* Outarr =NewBasevar[len]; //D-Bit loop     for(intj=0; j<d; ++j) { for(inti =0; i<len; ++i) {vararr[i].var= (intarr[i]>>d) &1;//take the value of each bitVararr[i].idx = i;//index before the record is sorted        }        //Count SortCOUNTING_SORT_EX (Vararr,outarr,Ten, Len);  for(intI=0; i<len; ++i) {Intarr[outarr[i].idx]= Intarr[vararr[i].idx];//assigns values on the original index based on the new index        }    }        Delete[] Vararr; Delete[] outarr;}

Bucket sort

Assuming that the input data is uniformly distributed, the time cost for the average is O (n).

Main ideas:

The data is divided into n equal-sized intervals, called buckets. The input is then distributed as evenly as possible to each bucket by mapping function f (k). Finally, each bucket is sorted separately,

Then walk through each bucket and list the elements in each bucket in order.

Code slightly.

Reading diary-Linear time Sorting algorithm

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