五种内部排序C语言实现

1.插入排序：

bool Insert(List L, ElementType X) {
	if (L == NULL)return false;
	if (L->Last == MAXSIZE - 1)return false;
	for (int i = 0; i <= L->Last; i++)
		if (L->Data[i] == X)return false;
	int i = 0;
	for(;i<L->Last+1;i++)
		if (X > L->Data[i]) {
			for (int j = L->Last; j >= i; j--)
				L->Data[j + 1] = L->Data[j];
			break;
		}
		else if (X == L->Data[i])return false;
		L->Data[i] = X;
		L->Last++;
	return true;
}

2.选择排序（快排）

int partition(SqList &L,int low,int hight){
    
    int pivotkey = L.r[low].key;
    while(low<high){
        while(low<high&&L.r[high].key>=pivotkey)  --high;
        
        swap(L.r[low],L.r[high]);

        while(low<high&&L.r[low].key<=pivotkey) ++low;
        
        swap(L.r[low],L.r[high]);
    
    }
    return low;

}

void Qsort(SqList &L,int low ,int high){
    
     if(low<high){
    
    pivotlc = partition(L,low,high);
    Qsort(L,low,pivotlc-1);
    Qsort(L,pivolc+1,high);
    }
}

3.选择排序（堆排）

#include <stdio.h>
#include <stdlib.h>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<math.h>
#define MAXSIZE 10
 
typedef struct MyStruct
{
	int arr[10]={-1,8,7,1,5,3,9,2,4,6};
	int size = MAXSIZE;
}heap;
heap h1;
void adjust(int last) {
	for (int i = last; i >= 1; i--) {
		int j = i;
		int temp = h1.arr[i];
		for (; j >= 2 && h1.arr[j/2] < h1.arr[j]; j /= 2)
			h1.arr[j] = h1.arr[j/2];
		h1.arr[j] = temp;		
		for (int i = 0; i <MAXSIZE; i++)
			printf("%d,", h1.arr[i]);
		printf("\n");
	}
		
}
int main() {

	adjust(MAXSIZE-1);
	printf("break\n");
	for (int i = MAXSIZE - 1; i >= 1; i--) {
		int t = h1.arr[i];
		h1.arr[i] = h1.arr[1];
		h1.arr[1] = t;
		adjust(i-1);
		printf("num %d adj finished\n", MAXSIZE - i);
	}
	for (int i = 0; i <MAXSIZE; i++)
		printf("%d,", h1.arr[i]);

	return 0;
}

4.归并排序

#include <stdio.h>
 
#define ElementType int
#define MAXN 100
 
void merge_pass( ElementType list[], ElementType sorted[], int N, int length );
 
void output( ElementType list[], int N )
{
    int i;
    for (i=0; i<N; i++) printf("%d ", list[i]);
    printf("\n");
}
 
void  merge_sort( ElementType list[],  int N )
{
    ElementType extra[MAXN];  /* the extra space required */
    int  length = 1;  /* current length of sublist being merged */
    while( length < N ) { 
        merge_pass( list, extra, N, length ); /* merge list into extra */
        output( extra, N );
        length *= 2;
        merge_pass( extra, list, N, length ); /* merge extra back to list */
        output( list, N );
        length *= 2;
    }
} 
 
 
int main()
{
    int N, i;
    ElementType A[MAXN];
 
    scanf("%d", &N);
    for (i=0; i<N; i++) scanf("%d", &A[i]);
    merge_sort(A, N);
    output(A, N);
 
    return 0;
}




void merge_pass(ElementType list[], ElementType sorted[], int N, int length) {
    int temp=0;
    int pos = 0;
        while (pos < N) {
            int left = pos;
            int right = pos + length;
            while (left < pos + length&&right < pos + 2 * length&&left < N&&right < N) {
                if (list[left] < list[right])
                    sorted[temp++] = list[left++];
                else 
                    sorted[temp++] = list[right++];
            }
            while (left < pos + length&&left < N)
                sorted[temp++] = list[left++];
            while (right < pos + 2 * length&&right < N)
                sorted[temp++] = list[right++];
            pos = pos + length * 2;
        }}

5.

#include<iostream>
#include<vector>
#include<string>
using namespace std;


//低位优先
//测试数据   278 109 63 930 589 184 505 269 8 83
int main() {

    int n = 0;
    int max = 0;
    cout << "输入排序元素的个数：" << endl;
    cin >> n;
    vector<int>input(n);
    cout << "输入n个元素：" << endl;
    for (int i=0; i < n; i++) {
        cin >> input[i];
        if (input[i] > max)
            max = input[i];
    }
    string s = to_string(max);
    
    int flag = 1;
    for (int i = 0; i < s.length(); i++) {
        vector<vector<int>>bucket(10);
        for (int j = 0; j < n; j++) {
            int tmp = input[j];
            tmp = (tmp / flag) % 10;
            bucket[tmp].push_back(input[j]);
        }
        int t = 0;
        for (int j = 0; j < 10; j++) {
            for (int k = 0; k < bucket[j].size(); k++)
                input[t++] = bucket[j][k];
        }
        flag *= 10;
    }
    for (int i=0; i < n; i++)
        cout << input[i] << " ";
    return 0;
}