C Language algorithm

You can use the divide and conquer method to solve the problem of sorting, which is to arrange n elements in a non descending order. The divide-and-conquer approach usually uses the following steps to sort the algorithm: if n is 1, the algorithm terminates; otherwise, the set of elements is divided into two or more child sets, each of which is sorted separately, and the ordered subset is merged into a single set.

Suppose that only the set of n elements is divided into two child sets. Now you need to determine how to partition the Child collection. One possibility is to put the n-1 element in the first subset (called a) and the last element in the second set (called B). Sort a recursively by this way. Since b contains only one element, it is sorted, and after a row, just use the function i n s e r T in program 2-1 0 to combine a and B. Comparing this sort algorithm with I n s e r t i o n s o r t (see program 2-1 5), we can find that this sort algorithm is actually a recursive algorithm for insertion ordering. The complexity of the algorithm is O (n 2). Another way to divide n elements into two sub sets is to place the elements with the largest value in B and the rest in a. Then a is sorted recursively. To merge sorted A and B, just add B to a. If you use the function m a x (see program 1-3 1) to find the maximum element, this sort algorithm is actually a recursive algorithm for S e l e c t i o n s o r t (see program 2-7).

If you use the bubbling process (see program 2-8) to find the largest element and move it to the rightmost position, this sort algorithm is the recursive algorithm for B u b l e S o r t (see program 2-9). The complexity of the two recursive sorting algorithms is (N2). The complexity of the algorithm is O (n2) (see examples 2-1 6 and 2-1 7) if it is found to terminate the recursive division of a If a has been ordered.

The above partitioning scheme divides n elements into two very unbalanced sets a and B. A has n-1 elements, and B contains only one element. Here's a look at what happens with the balanced split method: A set contains n/k elements, and B contains the rest of the elements. Recursively use the divide-and-conquer method to sort a and B. Then a process called merge (M e rg e) is used to combine the ordered A and b into a single set.

Example 2-5 considers 8 elements, and the values are [1 0,4,6,3,8,2,5,7] respectively. If k = 2 is selected, then [1 0, 4, 6, 3] and [8, 2, 5, 7] are individually sorted separately. The results were [3, 4, 6, 1 0] and [2, 5, 7, 8]. The two sorted sequences are merged from the head of the two sequences. The element 2:3 is smaller, is moved to the result sequence, 3 is compared with 5, 3 is moved into the result sequence, 4 is compared to 5, 4 is put into the result sequence; 5 and 6 Compare, ... If you select K= 4, then the sequence [1 0, 4] and [6, 3, 8, 2, 5, 7] will be sorted. The results of the sorting were [4, 1 0] and [2, 3, 5, 6, 7, 8]. When these two sorted sequences are merged, the desired sequence is obtained.

Figure 2-6 shows the pseudo code for the divide and conquer sorting algorithm. The number of neutron sets in the algorithm contains n/k elements in the 2,a.

template
void sort( T E, int n)
{ / /对E中的n个元素进行排序，k为全局变量
if (n >= k) {
i = n/k;
j = n-i;
令A 包含E中的前i个元素
令B 包含E中余下的j个元素
s o r t ( A , i ) ;
s o r t ( B , j ) ;
m e rge(A,B,E,i,j,); //把A 和B 合并到E
}
else 使用插入排序算法对E 进行排序
}

Fig. 14-6 pseudo code of the divide and conquer sorting algorithm