内部排序算法4(归并排序)
2017-05-02 16:33
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二路归并排序
思想
将两个有序表合成一个新的有序表就是二路归并。例如,在元素序列L中有两个已经排好序的有序顺序表L[left],...,L[mid]和L[mid+1],...,L[right],它们可以归并成为一个有序表,仍然存放于L[left],...,L[right]中。在归并排序中,用变量i和j分别做L中两个表的当前检测指针,用变量k做归并后结果表L1的当前存放指针。当i和j都在两个表的表长内变化时,根据相应的两个元素的排序码大小,依次把排序码小的元素放到结果表中;当i与j中有一个已经超出表长时,将另一个表中的剩余部分照抄到结果表中。
算法实现
非递归算法
图片来自:http://www.cnblogs.com/crx234/p/5858488.html
//归并排序头文件 #pragma once #include<iostream> typedef int ElementType; class Merge { public: Merge(int length); void create(); void sort(); void print(); ~Merge(); private: ElementType *elem; int len; void mergeHelp(ElementType *&L1, ElementType *&L2, int left, int mid, int right); void mergepass(ElementType *&L1, ElementType *&L2, int l); }; Merge::Merge(int length) { len = length; elem = new ElementType[len]; } inline void Merge::create() { std::cout << "please input the list: " << std::endl; for (int i = 0; i < len; i++) { ElementType temp; std::cin >> temp; elem[i] = temp; } std::cout << "input successful" << std::endl; } inline void Merge::sort() //归并排序,不断的增大块的大小,从1开始,一直到大于等于len { int count = 1; ElementType *L1 = new ElementType[len]; while (count < len) { mergepass(elem, L1, count); count *= 2; mergepass(L1, elem, count); count *= 2; } } inline void Merge::print() { for (int i = 0; i < len; i++) { std::cout << elem[i] << " "; } std::cout << std::endl; } Merge::~Merge() { delete[] elem; } inline void Merge::mergeHelp(ElementType *&L1, ElementType *&L2, int left, int mid, int right) //两个小块进行合并,left是左块的开始,mid是两块的分界线,right是右块的尽头 { int i = left, j = mid + 1, k = left; while (i <= mid && j <= right) { if (L1[i] <= L1[j]) { L2[k++] = L1[i++]; } else { L2[k++] = L1[j++]; } } while (i <= mid) { L2[k++] = L1[i++]; } while (j <= right) { L2[k++] = L1[j++]; } } inline void Merge::mergepass(ElementType *&L1, ElementType *&L2, int l) //l是归并排序中每个子排序表的初始长度 { int i; for (i = 0; i + 2 * l < len; i = i + 2 * l) { mergeHelp(L1, L2, i, i + l - 1, i + 2 * l - 1); } if (i + l <= len - 1) //如果第二个表长不足l { mergeHelp(L1, L2, i, i + l - 1, len - 1); } else //如果只剩下一个表,直接复制就好 { for (int j = i; j < len; j++) { L2[j] = L1[j]; } } }
//归并排序main文件 using namespace std; #include "MergeSort.h" int main() { Merge m(10); m.create(); m.sort(); m.print(); system("pause"); }
结果
递归实现
图示
图片来自:百度百科
//递归归并排序头文件 #pragma once #include<iostream> typedef int ElementType; class Merge { public: Merge(int length); void create(); void print(); void recursiveSort(); ~Merge(); private: ElementType *elem; int len; void mergeHelp(ElementType *&L1, ElementType *&L2, int left, int mid, int right); void recursiveSortHelp (int left, int right); }; Merge::Merge(int length) { len = length; elem = new ElementType[len]; } inline void Merge::create() { std::cout << "please input the list: " << std::endl; for (int i = 0; i < len; i++) { ElementType temp; std::cin >> temp; elem[i] = temp; } std::cout << "input successful" << std::endl; } inline void Merge::print() { for (int i = 0; i < len; i++) { std::cout << elem[i] << " "; } std::cout << std::endl; } inline void Merge::recursiveSort() { recursiveSortHelp(0, len - 1); } inline void Merge::recursiveSortHelp(int left, int right) //分治法 { ElementType *temp = new ElementType[len]; if (left < right) { int mid = (left + right) / 2; recursiveSortHelp(left, mid); //处理左子串 recursiveSortHelp(mid + 1, right); //处理右子串 mergeHelp(elem, temp, left, mid, right); for (int i = left; i <= right; i++) { elem[i] = temp[i]; } } } Merge::~Merge() { delete[] elem; } inline void Merge::mergeHelp(ElementType *&L1, ElementType *&L2, int left, int mid, int right) //每两个小块进行合并 { int i = left, j = mid + 1, k = left; while (i <= mid && j <= right) { if (L1[i] <= L1[j]) { L2[k++] = L1[i++]; } else { L2[k++] = L1[j++]; } } while (i <= mid) { L2[k++] = L1[i++]; } while (j <= right) { L2[k++] = L1[j++]; } }
//递归归并排序main文件,采用分治法 using namespace std; #include "MergeSort.h" int main() { Merge m(10); m.create(); m.recursiveSort(); m.print(); system("pause"); }
结果
算法分析
时间复杂度
MergePass做一趟二路归并排序调用MergeHelp函数2ln次,其中l是需要归并子序列的长度,n是整个序列的长度。而MergeSort调用MergePass正好⌈log2n⌉次,而每次做MergeHelp要执行比较O(l)次,所以算法总的排序码比较次数为O(nlog2n)。不论元素初始排列如何买这个时间的代价是不变的。空间复杂度
使用了一个与原待排序元素数组相同大小的辅助数组,空间代价为O(n)。算法的稳定
归并排序是稳定的。注:本文参考书籍《数据结构精讲与习题详解—考研辅导与答疑解惑》,殷人昆编著,清华大学出版社。