归并排序

来源：互联网发布：简单的java代码编辑：程序博客网时间：2024/06/06 12:32

采用分治的思想以O(NlogN)最坏的情形运行时间运行

如果对merge的每个递归调用都采用局部声明一个临时数组，那么在任一时刻就可能有logN个临时数组处在活动期

代码如下：

 1 #include <iostream> 2 #include <vector> 3 using namespace std; 4 template <typename Comparable> 5 void mergeSort(vector<Comparable> & a) 6 { 7     vector<Comparable> tmpArray(a.size()); 8     mergeSort(a,tmpArray,0,a.size()-1); 9 }10 template <typename Comparable>11 void mergeSort(vector<Comparable> & a,vector<Comparable> & tmpArray,int left,int right)12 {13     if(left<right)14     {15         int center = (left+right)/2;16         mergeSort(a,tmpArray,left,center);17         mergeSort(a,tmpArray,center+1,right);18         merge(a,tmpArray,left,center+1,right);19     }20 }21 template <typename Comparable>22 void merge(vector<Comparable> & a,vector<Comparable> & tmpArray,int leftPos,int rightPos,int rightEnd)23 {24     int leftEnd = rightPos-1;25     int tmpPos = leftPos;26     int numElements = rightEnd - leftPos + 1;27 28     while(leftPos <= leftEnd && rightPos <= rightEnd)29         if(a[leftPos] <= a[rightPos])30             tmpArray[tmpPos++] = a[leftPos++];31         else32             tmpArray[tmpPos++] = a[rightPos++];33 34     while(leftPos <= leftEnd)35         tmpArray[tmpPos++] = a[leftPos++];36     while(rightPos <= rightEnd)37         tmpArray[tmpPos++] = a[rightPos++];38 39     for(int i=0;i<numElements;i++,rightEnd--)40         a[rightEnd] = tmpArray[rightEnd];41 }42 int main()43 {44     vector<int> ivec;45     ivec.push_back(1);46     ivec.push_back(9);47     ivec.push_back(2);48     ivec.push_back(10);49     ivec.push_back(3);50     ivec.push_back(11);51     ivec.push_back(4);52     ivec.push_back(12);53     ivec.push_back(5);54     ivec.push_back(13);55     ivec.push_back(6);56     ivec.push_back(14);57     ivec.push_back(7);58     ivec.push_back(15);59     ivec.push_back(8);60     ivec.push_back(16);61     mergeSort(ivec);62     for(int j = 0;j<ivec.size();j++)63         cout<<ivec[j]<<endl;64     return 0;65 }

运行结果：

阅读全文

0 0