C++ program to implement quick sort algorithm #include <iostream>usingnamespacestd;voidquicksort(int,int,int);intpartition(int,int,int);intpartition(int*a,ints,inte) {intpiviot=a[e];intpind=s;inti, t;for(i=s; i<e; i++) {if(a[i]<=piviot) { t=a[i]; a[i]=a[pind]; a[...
intmain(){intsize;printf("Enter the size of array : ");scanf("%d",&size);intarr[size],i;for(i=0;i<size;i++)scanf("%d",&arr[i]);quick_sort(arr,0,size-1);for(i=0;i<size;i++)printf("%d\t",arr[i]);printf("\n");return0; } 开发者ID:abhisheknith,项目名称:Code,代码...
template<typename T>voidquick_sort_T(T *arr,intlow,inthigh) {if(low<high) {intpivot=get_partition(arr,low,high); quick_sort_T(arr,low,pivot-1); quick_sort_T(arr,pivot+1,high); } } void quick_sort_len(const int &len) { uint32_t *arr = new uint32_t[len]; gen_T_array_...
通常来说,为了避免快速排序退化为冒泡排序,以及递归栈过深的问题,我们一般依据“三者取中”的法则来选取基准元素,“三者”即序列首元素、序列尾元素、序列中间元素,在三者中取中值作为本趟快速排序的基准元素。 原文链接:图解快排--快速排序算法(quick sort) ...
long long mergeSort(int d[],int len){ if(len==1){ return 0; } int l = len/2; long long lNum = mergeSort(d,l); //左边部分的逆序对数 long long rNum = mergeSort(d+l,len-l); //右边部分的逆序对数 sort(d,d+l);
Quicksort是一个很好的排序算法,但是其最坏情况运行时间是O(n^2), 还不如Mergesort的O(nlgn),如何改进Quicksort? 引进随机化思想一种方法: 对给定的待排序序列,随机地重排列另一种方法:随机选取pivot给出第二种方法的代码1/** 2 * 2010.1.24 3 * 随机快速排序法 4 * pivot元素并不是简单地取第一个元...
Here You will find solutions to various DSA problems. These are standard questions published on different platform like leetcode, codeforces etc. - Solving-DSA-Problems/QuickSort.cpp at main · ankit-0369/Solving-DSA-Problems
The performance of CPP11sort is evaluated and compared with its mainstream competitors provided by GNU, Intel, and Microsoft. It is shown that out of the considered implementations, CPP11sort mostly yields the shortest sorting times and is the only one that is portable to any conforming C++ ...
// main.cpp // greedy #include <iostream> using std::cout; using std::cin; using std::string; #define SIZEOF_ARRAY(a) (sizeof(a)/sizeof(a[0])) template<typename T> void insertion_sort(T *a, size_t n) { T tmp; size_t j, p; for (p = 1; p < n; p++) { tmp = ...
voidmerge(int*a,intl,intmid,intr) { inti,j,k; i=0,j=l,k=mid; while(j<mid&&k<r) { if(a[j]>a[k]) { sum+=mid-j; temp[i++]=a[k++]; } elsetemp[i++]=a[j++]; } while(j<mid) temp[i++]=a[j++]; while(k<r) ...