Data Structure Tutorial 36th Lesson Select Sort, Merge sort

Teaching Purpose: Mastering the choice of sorting, merging the sorting algorithm

Teaching emphases: Choosing the sort of heap and merging sort algorithm

Teaching Difficulty: heap Sort algorithm

Teaching Content:

One, select the sort

Each trip in the n-i+1 (i=1,2,... n-1) record selects the record with the smallest keyword as the first record in the ordered sequence.

Second, the simple choice of sorting

Algorithm:

Smp_selecpass (ListType &r,int i)

{

K=i;

for (j=i+1;j<n;i++)

if (R[j].key<r[k].key)

K=j;

if (k!=i)

{t=r[i];r[i]=r[k];r[k]=t;}

}

Smp_sort (ListType &r)

{

for (i=1;i<n-1;i++)

Smp_selecpass (R,i);

}

Three, tree-type selection sorting

Also called tournament sort, the keywords for n records are compared to 22, and then 22 comparisons are made between half of the smaller ones, so repeat until a record of the minimum key is selected.

Iv. Heap Sequencing

Only one secondary space is required for the record size, and each record to be sorted occupies only one storage space.

What is a heap? The sequence of n elements {k1,k2,..., kn} is called a heap if and only if the following relationships are satisfied. Relationship One: ki<=k2i Relationship II: ki<=k2i+1 (i=1,2,..., N/2)

Heap sorting solves two problems: 1, how to build a heap from a unordered sequence? 2. How do I adjust the remaining elements to become a new heap after I output the top element of the heap?

Solution to question 2:

Sift (ListType &r,int k,int m)

{

I=k;j=2*i;x=r[k].key;finished=false;

T=R[K];

while ((j<=m) && (!finished))

{

if ((j<m) && (R[j].key>r[j+1].key)) J + +;

if (X<=r[j].key)

Finished:=true;

else {r[i]=r[j];i=j;j=2*i;}

}

r[i]=t;

}

Heapsort (ListType &r)

{

for (i=n/2;i>0;i--) sift (r,i,n);

for (i=n;i>1;i--) {

r[1]<->r[i];

Sift (r,i,i-1)

}

}

Five, merge sort

The method of combining two or more ordered tables into a new ordered table is called merge.

Assuming that the initial sequence contains n records, it is considered to be N ordered subsequence, with a length of 1 and then 22 merged to obtain an ordered sequence of N/2 with a length of 2 or 1, then 22 merge, so repeat.

Merge (ListType r,int l,int m,int n,listtype &r2)

{

P>I=L;J=M+1;K=L-1;

while (i<=m) and (j<n) do

{

k=k+1;

if (r[i].key<=r[j].key) {r2[k]=r[i];i++;}

Else {r2[i]=r[j];j++}

}

if (i<=m) r2[k+1..n]=r[i ... M];

if (j<=n) r2[k+1..n]=r[j ... n];

}

MergeSort (listtype &r,listtype &r1,int s,int t)

{

if (s==t)

R1[s]=r[s];

Else

{

MergeSort (R,R2,S,S+T/2);

MergeSort (r,r2,s+t/2+1,t);

Merge (R2,S,S+T/2,T,R1);

}

]

Vi. Summary