首先是算法实现文件Sort.h，代码如下：
/** 实现了八个常用的排序算法：插入排序、冒泡排序、选择排序、希尔排序* 以及快速排序、归并排序、堆排序和LST基数排序* @author gkh178*/#include <iostream>template<class T>void swap_value(T &a, T &b){T temp = a;a = b;b = temp;}//插入排序：时间复杂度o(n^2)template<class T>void insert_sort(T a[], int n){for (int i = 1; i < n; ++i){T temp = a[i];int j = i – 1;while (j >= 0 && a[j] > temp){a[j + 1] = a[j];–j;}a[j + 1] = temp;}}//冒泡排序：时间复杂度o(n^2) template<class T>void bubble_sort(T a[], int n){for (int i = n – 1; i > 0; –i){for (int j = 0; j < i; ++j){if (a[j] > a[j + 1]){swap_value(a[j], a[j + 1]);}}}}//选择排序：时间复杂度o(n^2)template<class T>void select_sort(T a[], int n){for (int i = 0; i < n – 1; ++i){T min = a[i];int index = i;for (int j = i + 1; j < n; ++j){if (a[j] < min){min = a[j];index = j;}}a[index] = a[i];a[i] = min;}}//希尔排序：时间复杂度介于o(n^2)和o(nlgn)之间 template<class T>void shell_sort(T a[], int n){for (int gap = n / 2; gap >= 1; gap /= 2){for (int i = gap; i < n; ++i){T temp = a[i];int j = i – gap;while (j >= 0 && a[j] > temp){a[j + gap] = a[j];j -= gap;}a[j + gap] = temp;}}}//快速排序：时间复杂度o(nlgn) template<class T>void quick_sort(T a[], int n){_quick_sort(a, 0, n – 1);}template<class T>void _quick_sort(T a[], int left, int right){if (left < right){int q = _partition(a, left, right);_quick_sort(a, left, q – 1);_quick_sort(a, q + 1, right);}}template<class T>int _partition(T a[], int left, int right){T pivot = a[left];while (left < right){while (left < right && a[right] >= pivot){–right;}a[left] = a[right];while (left < right && a[left] <= pivot){++left;}a[right] = a[left];}a[left] = pivot;return left;}//归并排序：时间复杂度o(nlgn) template<class T>void merge_sort(T a[], int n){_merge_sort(a, 0, n – 1);}template<class T>void _merge_sort(T a[], int left, int right){if (left < right){int mid = left + (right – left) / 2;_merge_sort(a, left, mid);_merge_sort(a, mid + 1, right);_merge(a, left, mid, right);}}template<class T>void _merge(T a[], int left, int mid, int right){int length = right – left + 1;T *newA = new T[length];for (int i = 0, j = left; i <= length – 1; ++i, ++j){*(newA + i) = a[j];}int i = 0;int j = mid – left + 1;int k = left;for (; i <= mid – left && j <= length – 1; ++k){if (*(newA + i) < *(newA + j)){a[k] = *(newA + i);++i;}else{a[k] = *(newA + j);++j;}}while (i <= mid – left){a[k++] = *(newA + i);++i;}while (j <= right – left){a[k++] = *(newA + j);++j;}delete newA;}//堆排序：时间复杂度o(nlgn) template<class T>void heap_sort(T a[], int n){built_max_heap(a, n);//建立初始大根堆 //交换首尾元素，并对交换后排除尾元素的数组进行一次上调整 for (int i = n – 1; i >= 1; –i){swap_value(a[0], a[i]);up_adjust(a, i);}}//建立一个长度为n的大根堆 template<class T>void built_max_heap(T a[], int n){up_adjust(a, n);}//对长度为n的数组进行一次上调整 template<class T>void up_adjust(T a[], int n){//对每个带有子女节点的元素遍历处理，从后到根节点位置 for (int i = n / 2; i >= 1; –i){adjust_node(a, n, i);}}//调整序号为i的节点的值 template<class T>void adjust_node(T a[], int n, int i){//节点有左右孩子 if (2 * i + 1 <= n){//右孩子的值大于节点的值，交换它们if (a[2 * i] > a[i – 1]){swap_value(a[2 * i], a[i – 1]);}//左孩子的值大于节点的值，交换它们if (a[2 * i – 1] > a[i – 1]){swap_value(a[2 * i – 1], a[i – 1]);}//对节点的左右孩子的根节点进行调整adjust_node(a, n, 2 * i);adjust_node(a, n, 2 * i + 1);}//节点只有左孩子，为最后一个有左右孩子的节点 else if (2 * i == n){//左孩子的值大于节点的值，，交换它们if (a[2 * i – 1] > a[i – 1]){swap_value(a[2 * i – 1], a[i – 1]);}}}//基数排序的时间复杂度为o(distance(n+radix)),distance为位数，n为数组个数，radix为基数 //本方法是用LST方法进行基数排序，MST方法不包含在内 //其中参数radix为基数，一般为10；distance表示待排序的数组的数字最长的位数；n为数组的长度 template<class T>void lst_radix_sort(T a[], int n, int radix, int distance){T* newA = new T[n];//用于暂存数组 int* count = new int[radix];//用于计数排序，保存的是当前位的值为0 到 radix-1的元素出现的的个数 int divide = 1;//从倒数第一位处理到第一位 for (int i = 0; i < distance; ++i){//待排数组拷贝到newA数组中for (int j = 0; j < n; ++j){*(newA + j) = a[j];}//将计数数组置0for (int j = 0; j < radix; ++j){*(count + j) = 0;}for (int j = 0; j < n; ++j){int radixKey = (*(newA + j) / divide) % radix; //得到数组元素的当前处理位的值(*(count + radixKey))++;}//此时count[]中每个元素保存的是radixKey位出现的次数//计算每个radixKey在数组中的结束位置，位置序号范围为1-nfor (int j = 1; j < radix; ++j){*(count + j) = *(count + j) + *(count + j – 1);}//运用计数排序的原理实现一次排序，排序后的数组输出到a[]for (int j = n – 1; j >= 0; –j){int radixKey = (*(newA + j) / divide) % radix;a[*(count + radixKey) – 1] = newA[j];–(*(count + radixKey));}divide = divide * radix;}}然后是测试文件main.cpp，代码如下：
在前进的路上，主动搬开别人脚下的绊脚石，有时往往也是为自己铺路。
