算法导论:主要关注的是程序的性能;速度令人渴望!!!
排序算法是经典算法
1、插入排序
(1)、算法模型
(2)、代码实现
#includevoid insertSort(int *a, int count); void showArray(int *a, int count); void showArray(int *a, int count){ int i; for(i = 0; i < count; i++){ printf("%d ", a[i]); } printf("\n"); } void insertSort(int *a, int count){ int i; int j; int n; int tmp; for(i = 1; i < count; i++){ tmp = a[i]; for(j = 0; a[i]>a[j] && j j; n--){ a[n] = a[n-1]; } a[j] = tmp; } } } void main(void){ int a[] = {2, 5, 7, 1, 11, 0, 6, 9}; int count = sizeof(a)/sizeof(int); printf("排序前输出如下: "); showArray(a, count); insertSort(a, count); printf("排序后输出如下: "); showArray(a, count); }
(3)、结果截图
(4)、算法分析
插入排序最坏的情况:数组中所有元素全部逆序排列;
时间复杂度:O(n^2);
2、归并排序
(1)、算法思想:
i、if n = 1; done
ii、递归排序,分2部分,在[0, n/2]和[n/2, n]
iii、将2部分归并排序
(2)、核心代码实现
#include#include void mergerSort(int *a1, int *a2, int **a3, int count1, int count2, int *count3); void showArray(int *a3, int count); void showArray(int *a3, int count){ int i; for(i = 0; i < count; i++){ printf("%d ", a3[i]); } printf("\n"); } void mergerSort(int *a1, int *a2, int **a3, int count1, int count2, int *count3){ int count; int i = 0; int j = 0; int n = 0; count = *count3 = count1 + count2; *a3 = (int *)malloc(sizeof(int) * count); //以下的都是<,因为传过来的是数组长度; while(i < count1 && j < count2){ if(a1[i] < a2[j]){ (*a3)[n++] = a1[i]; i++; }else if(a1[i] == a2[j]){ (*a3)[n++] = a1[i]; (*a3)[n++] = a2[j]; i++; j++; }else{ //刚才写程序else(a1[i] > a2[j]),后发现else语句后面是没有条件的!!! (*a3)[n++] = a2[j]; j++; } } while(i < count1){ (*a3)[n++] = a1[i]; i++; } while(j < count2){ (*a3)[n++] = a2[j]; j++; } } /* 归并排序核心算法就是:将已经排好序的2个数组进行最终的排序过程; */ void main(void){ int a1[] = {1, 3, 5, 7}; int a2[] = {0, 2, 4, 6, 8, 9, 10}; int count1 = sizeof(a1)/sizeof(int); int count2 = sizeof(a2)/sizeof(int); int *a3 = NULL; int count3 = 0; mergerSort(a1, a2, &a3, count1, count2, &count3); showArray(a3, count3); free(a3); }
(3)、结果截图
(4)、完整代码实现
#include#include void mergeSort(int *a, int low, int high); void merge(int *a, int low, int mid, int high); void merge(int *a, int low, int mid, int high){ int i = low; int j = mid+1; int n = 0; int *a2; a2 = (int *)malloc(sizeof(int) * (high-low+1)); if(a2 == NULL){ return; } //以下都是<=,因为传过来的都是下标; while(i <= mid && j <= high){ if(a[i] < a[j]){ a2[n++] = a[i]; i++; }else if(a[i] == a[j]){ a2[n++] = a[i]; i++; j++; }else{ a2[n++] = a[j]; j++; } } while(i <= mid){ a2[n++] = a[i]; i++; } while(j <= high){ a2[n++] = a[j]; j++; } for(n = 0, i = low; i <= high; n++, i++){ //将a2中的元素复制回a中; a[i] = a2[n]; } free(a2); } void mergeSort(int *a, int low, int high){ int mid; if(low < high){ mid = (low + high) / 2; mergeSort(a, low, mid); mergeSort(a, mid+1, high); merge(a, low, mid, high); } } void main(void){ int a[] = {2, 4, 1, 7, 5, 6, 9, 10}; int count = sizeof(a)/sizeof(int); int i; mergeSort(a, 0, count-1); for(i = 0; i < count; i++){ printf("%d ", a[i]); } printf("\n"); }
(5)、结果截图
(6)、算法分析
归并排序的时间复杂度:树高度log(n),一共要对n个元素进行排序,所以为:O(nlogn);
在30个元素以内,插入排序性能更好,超过30个元素之后归并排序的性能更加优秀;