排列组合，遍历抽样

以前只碰过已知子样本大小的抽样。如从5个数抽出所有的2个数的组合，用2个for循环就可以搞定了。

A01  A12  A23  A34

A02  A13  A24

A03  A14 

A04

 

从5个数抽出所有的3个数的组合，用3个for循环就可以搞定了。

A012  A123  A234

A013  A124

A014

A023  A134

A024

A034

 

从N个数中抽取所有n个数的组合呢？

 

N = sampleSize

subSampleSize = n

 

 

struct Sample{int subSample[10];};// 遍历采样bool Sampling(vector<Sample> &samplingResult, int sampleSize, int subSampleSize){if(sampleSize<=1)return false;const int MAX_SUB_SAMPLE_SIZE = 10;// 最大子样本维数vector<Sample> vecSampleing[MAX_SUB_SAMPLE_SIZE];for(int i=0; i<subSampleSize; i++){if(i==0){for(int j=0; j<sampleSize; j++){Sample temp;temp.subSample[i] = j;vecSampleing[i].push_back( temp );}}else{int lastSampleLevelNum = int(vecSampleing[i-1].size());for(int p=0; p<lastSampleLevelNum; p++){for(int j=0; j<sampleSize; j++){if(j>vecSampleing[i-1][p].subSample[i-1]){Sample temp;for(int q=0; q<i; q++){temp.subSample[q] = vecSampleing[i-1][p].subSample[q];}temp.subSample[i] = j;vecSampleing[i].push_back(temp);}}}}}int sampleNumTotal = vecSampleing[subSampleSize-1].size();for(int i=0; i<sampleNumTotal; i++){samplingResult.push_back( vecSampleing[subSampleSize-1][i] );}return true;}