//使用i、j、k代表甲队,分别匹配乙队var team2=['x','y','z'];for(var i =0; i <3; i++){for(var j =0; j <3; j++){for(var k =0; k <3; k++){//按照题目要求写判断if(j != k && i != k && i != j && k !=0&& k !=2&& i !=0){
console.log('a'+ team2[i]+'\n'+'b'+ team2[j]+'\n'+'c'+ team2[k]+'\n');}}}}
下面这个是我最开始写的一个“失败品”,虽然结果正确,但是过程太复杂,大家看看就好
var team1 =['a','b','c'];var team2 =['x','y','z'];var all =[];while(all.length <3){for(var i =0; i < team1.length; i++){var oneArr =[];for(var j =0; j < team2.length; j++){if(team1[i]=='a'&& team2[j]=='x'){continue;}elseif((team1[i]=='c'&& team2[j]=='x')||(team1[i]=='c'&& team2[j]=='z')){continue;}else{var str = team1[i]+ team2[j];
oneArr.push(str);}}
all.push(oneArr);}for(var x =0; x < all.length; x++){//判断是否为object,也就是数组if(typeof(all[x])=='object'){//如果是数组,并且长度为1,则证明这个是唯一的,把他提取出来if(all[x].length ==1){
all[x]= all[x][0];for(var t1 =0; t1 < team1.length; t1++){if(team1[t1]== all[x][0]){//从team1中,把这个元素删除掉
team1.splice(t1,1);}}for(var t2 =0; t2 < team2.length; t2++){if(team2[t2]== all[x][1]){//从team2中,把这个元素删除掉
team2.splice(t2,1);}}}}}for(i in all){if(typeof(all[i])=='object'){
all.splice(i,1);}}}
console.log(all);