TA貢獻1812條經驗 獲得超5個贊
遞歸publicListpermutation(List inputList){ ListresList=newArrayList<>(); permutationInt(inputList,resList,0,newchar[inputList.size()]);returnresList;}privatevoidpermutationInt(ListinputList,List resList, intind,char[]arr){if(ind==inputList.size()){resList.add(newString(arr));return;}for(charc:inputList.get(ind).toCharArray()){arr[ind]=c;permutationInt(inputList,resList,ind+1,arr);}}
TA貢獻1854條經驗 獲得超8個贊
我知道你問的是java的，不過我給個python的，python比較方便嘿嘿，java的原理是一樣的L1=['a','b','c','d']L2=['1','2']L3=['g']L4=['q','w','e']list=[[i,j,k,p]foriinL1forjinL2forkinL3forpinL4]補充回答：不確定幾個list，不確定list多長的時候#把取來的數組合成一個大list，然后再一行代碼就沒有然后了s=[['a','b','c','d'],['1','2'],['g'],['q','w','e']]list=list(itertools.product(*s))
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