謝謝你的文章。
根據(jù)你的思路， 寫了一個(gè)簡(jiǎn)單的算法，好像不需要上邊那么復(fù)雜。這里我用true代表1， false代表0?？纯从袥](méi)有問(wèn)題。
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public class TestFastCombination {

/**
* @param args
*/
public static void main(String[] args) {
TestFastCombination comb = new TestFastCombination();
comb.combine(5);
}


private void combine(int totalSize){
//int currentSize = 0;
boolean[] myList = new boolean[totalSize];
for(int currentSize = 3; currentSize<totalSize-1;currentSize++){
//initial
for(int i=0;i<totalSize;i++){
if(i<currentSize){
myList[i] = true;
}
else{
myList[i] = false;
}
}
//PrintFirst combination
for(int i = 0;i<myList.length ;i++){
System.out.print(myList[i]+" ");
}
System.out.println();

//start to search 10 and move left 1s to left most
findAndChange10(myList);
}


}

private void findAndChange10(boolean[] tempList){
int leftOneNumbers = 0;
int leftZeroNumbers = 0;
boolean isFirstPairMatch = false;
do{
leftOneNumbers = leftZeroNumbers = 0;
isFirstPairMatch = false;
for(int i=0;i<tempList.length-1 ; i++){

if(tempList[i]==true ){
leftOneNumbers++;
if(tempList[i+1]==false){
tempList[i]=false;
tempList[i+1]=true;
leftOneNumbers--;
if(i==0){
isFirstPairMatch = true;
}
break;
}
}
else{
leftZeroNumbers++;
}
}

if(leftOneNumbers>0){
for(int j=0;j<leftOneNumbers;j++){
tempList[j]=true;
}
if(leftZeroNumbers>0){
for(int j=0;j<leftOneNumbers;j++){
tempList[leftOneNumbers+j]=false;
}
}
}
//Here I got a other possible combination
for(int i = 0;i<tempList.length ;i++){
System.out.print(tempList[i]+" ");
}
System.out.println();
}
while(leftOneNumbers>0 ||(isFirstPairMatch));



}


}

對(duì)于C(5,3), 打印結(jié)果如下：
true true true false false
true true false true false
true false true true false
false true true true false
true true false false true
true false true false true
false true true false true
true false false true true
false true false true true
false false true true true