不，您不需要搜索所有可能的替代方法。一个简单的递归算法（如@recursive给出的算法）将为您提供答案。如果您正在寻找一个实际上输出所有组合而不是仅仅输出所有组合的函数，那么这里是用R编写的版本。我不知道您使用的是哪种语言，但是翻译它应该非常简单R擅长于这种事情，尽管代码可能会更长一些。
allCombos<-function(len, ## number of items to sample x, ## array of quantities of balls, by color names=1:length(x) ## names of the colors (defaults to "1","2",...)){ if(length(x)==0) return(c()) r<-c() for(i in max(0,len-sum(x[-1])):min(x[1],len)) r<-rbind(r,cbind(i,allCombos(len-i,x[-1]))) colnames(r)<-names r}这是输出：
> allCombos(3,c(3,3),c("white","black")) white black[1,] 0 3[2,] 1 2[3,] 2 1[4,] 3 0> allCombos(10,c(15,1,1,1,1,1),c("white","black","blue","red","yellow","green")) white black blue red yellow green [1,] 5 1 1 1 1 1 [2,] 6 0 1 1 1 1 [3,] 6 1 0 1 1 1 [4,] 6 1 1 0 1 1 [5,] 6 1 1 1 0 1 [6,] 6 1 1 1 1 0 [7,] 7 0 0 1 1 1 [8,] 7 0 1 0 1 1 [9,] 7 0 1 1 0 1[10,] 7 0 1 1 1 0[11,] 7 1 0 0 1 1[12,] 7 1 0 1 0 1[13,] 7 1 0 1 1 0[14,] 7 1 1 0 0 1[15,] 7 1 1 0 1 0[16,] 7 1 1 1 0 0[17,] 8 0 0 0 1 1[18,] 8 0 0 1 0 1[19,] 8 0 0 1 1 0[20,] 8 0 1 0 0 1[21,] 8 0 1 0 1 0[22,] 8 0 1 1 0 0[23,] 8 1 0 0 0 1[24,] 8 1 0 0 1 0[25,] 8 1 0 1 0 0[26,] 8 1 1 0 0 0[27,] 9 0 0 0 0 1[28,] 9 0 0 0 1 0[29,] 9 0 0 1 0 0[30,] 9 0 1 0 0 0[31,] 9 1 0 0 0 0[32,] 10 0 0 0 0 0>
