快速排序

单向扫描

//单向的void qsort1(int *arr, int begin, int end){if(begin >= end)return ;int m = begin;//Assert arr[begin+1....m] < arr[begin] && arr[m+1....i-1] >=arr[begin] )for(int i=begin+1; i<=end; i++){if(arr[i] < arr[begin]){//Swap(arr, ++m, i);m++;int tmp = arr[m];arr[m] = arr[i];arr[i] = tmp;}}//Assert arr[m] < arr[begin] && arr[m+1....end] >= arr[begin] //swap(arr, begin, m);int tmp = arr[begin];arr[begin] = arr[m];arr[m] = tmp;qsort1(arr, begin, m-1);qsort1(arr, m+1, end);}

单向扫描的劣势：针对数组所有元素相同情况下，交换次数太多，时间复杂度太高O(n^2)

 

 

双向扫描，在数组所有元素相同情况下，交换次数比qsort1少

//双向划分的快排void qsort2(int *arr, int begin, int end){ if(begin >= end) return ; int b = begin; int e = end; int key = arr[begin]; //arr[begin] as a key while(b < e) { while(b < end && arr[b] <= key) b++; while(e > begin && arr[e] >= key) e--; if(b < e) { int tmp = arr[b]; arr[b] = arr[e]; arr[e] = tmp; } } arr[begin] = arr[e]; arr[e] = key; qsort2(arr, begin, e - 1); qsort2(arr, e + 1, end);}

 

第二种做法，类似上面代码

但是感觉应该比较次数能减少一些

//双向划分的快排void qsort2(int *arr, int begin, int end){ if(begin >= end) return ; int b = begin; int e = end + 1; int key = arr[begin]; //arr[begin] as a key while(b < e) { while(b++ < e&& arr[b] <= key) ; while(e-- > b && arr[e] >= key) ; if(b < e) { int tmp = arr[b]; arr[b] = arr[e]; arr[e] = tmp; } } arr[begin] = arr[e]; arr[e] = key; qsort2(arr, begin, e - 1); qsort2(arr, e + 1, end);}

 

 

 

自动生成测试用例测试上述快排是否正确

随机打乱数组中的数

然后快排后，判断其是否有序

bool isSort(int *arr, int len){if(len <= 1)return true;int i = 0;while(i < len - 1){if(arr[i] > arr[i+1])return false;elsei++;}return true;}int main(){int arr[100];//Testfor(int j=0; j<100; j++){ //initfor(int i=0; i<100; i++)arr[i] = i; //打乱数组中元素for(int i=0; i<99; i++){int index = rand()%(100-i) + i;int tmp = arr[index];arr[index] = arr[i];arr[i] = tmp;}qsort2(arr, 0, 99);if(isSort(arr, 100));elseprintf("False/n");}printf("END/n"); system("pause"); return 0;}