09-排序1 排序

#include <stdio.h>#include <stdlib.h>void Bubble_Sort(long long int A[], int N);void Insertion_Sort(long long int A[], int N);void Shell_Sort(long long int A[], int N);void Heap_Sort(long long int A[], int N);void PercDown(long long int A[], int i, int N);void Merge_Sort(long long int A[], int N);void MSort(long long int A[], long long int tmpA[], int L, int RightEnd);void Merge_Sort_2(long long int A[], int N);void Merge_pass(long long int A[], long long int tmpA[], int N, int length);void Merge(long long int A[], long long int tmpA[], int L, int R, int RightEnd);void swap(long long int *a, long long int *b){long long int tmp;tmp = *a;*a = *b;*b = tmp;}int main(int argc, char const *argv[]){freopen("test.txt", "r", stdin);int N;scanf("%d", &N);long long int A[N], tmp;for (int i = 0; i < N; i++){scanf("%lld", &tmp);A[i] = tmp;}// Bubble_Sort(A, N);// Insertion_Sort(A, N);// Shell_Sort(A, N);Heap_Sort(A, N);// Merge_Sort(A, N);// Merge_Sort_2(A, N);for (int i = 0; i < N; i++){if (i == 0)printf("%d", A[i]);elseprintf(" %d", A[i]);}return 0;}void Bubble_Sort(long long int A[], int N){for (int P = N-1; P >= 0; P--){int flag = 0;for (int i = 0; i < P; i++){if (A[i] > A[i+1]){swap(&A[i], &A[i+1]);flag = 1;}}if (flag == 0)break; //no swap}}void Insertion_Sort(long long int A[], int N){for (int P = 1; P < N; P++){long long int tmp = A[P];int i;for (i = P; i > 0 && A[i-1] > tmp; i--)A[i] = A[i-1];A[i] = tmp;}}void Shell_Sort(long long int A[], int N){int D, P, i;long long int tmp;for (D = N/2; D > 0; D/=2){for (P = D; P < N; P++){tmp = A[P];for (i = P; i >= D && A[i-D] > tmp; i-=D)A[i] = A[i-D];A[i] = tmp;}}}void Heap_Sort(long long int A[], int N){/*要弄明白为什么是更新一半的元素。你可以在纸上画一画，对每一个元素从二叉树上从1开始标号。会发现标号为1 ， 2，..n/2 的结点刚好可以覆盖二叉树的所有路径，并且是从 n/2 到 1 去更新堆，这样的话就可以构成一个初始化的最大堆。每次更新最大堆时，都是沿着左右子树的路径一次更新，要左右子结点较大的元素往上移动，知道更新的元素大于左右子树的结点值。那么就成了一个新的最大堆。复杂度就是二叉树数的层数*/for (int i = N/2-1; i >= 0; i--)/* BuildHeap */  //这里是循环排成大根堆，建立初始堆; 因为从A[0]开始，所以N/2-1 PercDown(A, i, N);for (int i = N-1; i > 0; i--){swap(&A[0], &A[i]);PercDown(A, 0, i);}}void PercDown(long long int A[], int i, int N)//这个是排成大根推的函数，A[]是接收一个数组，i为根节点,N为节点总数{int Child;long long int tmp;for (tmp = A[i]; 2*i+1 < N; i = Child){Child = 2*i+1;if (Child < N-1 && A[Child+1] > A[Child])Child++;if (tmp < A[Child])A[i] = A[Child];elsebreak;}A[i] = tmp;}void Merge_Sort(long long int A[], int N)//递归算法{long long int *tmpA = (long long int*)malloc(sizeof(long long int)*N);if(tmpA != NULL){MSort(A, tmpA, 0, N-1);free(tmpA);}elsereturn;}void MSort(long long int A[], long long int tmpA[], int L, int RightEnd){int Center;if (L < RightEnd){Center = (L + RightEnd) / 2;MSort(A, tmpA, L, Center);MSort(A, tmpA, Center+1, RightEnd);Merge(A, tmpA, L, Center+1, RightEnd);}}void Merge_Sort_2(long long int A[], int N)//非递归算法{long long int *tmpA = (long long int*)malloc(sizeof(long long int)*N);if(tmpA != NULL){int length = 1;while (length < N){Merge_pass(A, tmpA, N, length);length *= 2;Merge_pass(tmpA, A, N, length);length *= 2;}free(tmpA);}elsereturn;}void Merge_pass(long long int A[], long long int tmpA[], int N, int length){int i;for (i = 0; i <= N-2*length; i += 2*length)Merge(A, tmpA, i, i+length, i+2*length-1);if (i+length < N)Merge(A, tmpA, i, i+length, N-1);elsefor (int j = i; j < N; j++)tmpA[j] = A[j];}void Merge(long long int A[], long long int tmpA[], int L, int R, int RightEnd){int LeftEnd = R - 1;int tmp = L;int Number = RightEnd - L + 1;while (L <= LeftEnd && R <= RightEnd){if (A[L] <= A[R])tmpA[tmp++] = A[L++];elsetmpA[tmp++] = A[R++];}while (L <= LeftEnd)tmpA[tmp++] = A[L++];while (R <= RightEnd)tmpA[tmp++] = A[R++];for(int i = 0; i < Number; i++, RightEnd--)//L, R have changedA[RightEnd] = tmpA[RightEnd];}