一元表的交集、相加、相减、打包元组

☜李長新☞

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import console; 
A = {1,2,3};
B = {5,3};
console.log("表A:")
console.dumpTable(A)
console.log("表B:")
console.dumpTable(B)
var 集合交集 = function(A,B){
	for(k,v in A){
		for(i=1;#B;1){
			if(table.find(A,B[i])=null){
			table.removeByValue(B,B[i])
			}
		}
	}
	return B; 
}
console.log("集合交集(A,B)：")
console.dumpTable(集合交集(A,B));
console.pause(true);

A = {1,2,3};
B = {5,3};
console.log("表A:")
console.dumpTable(A)
console.log("表B:")
console.dumpTable(B)
var 集合相加 = function(A,B){
	for(n=1;#B;1){
		if(table.find(A,B[n])=null){
			//console.log(B[n])
			table.push(A,B[n])
		}
	}
return A; 	
}
console.log("集合相加(A,B)：")
console.dumpTable(集合相加(A,B));
console.pause(true);

A = {1,2,3};
B = {5,3};
console.log("表A:")
console.dumpTable(A)
console.log("表B:")
console.dumpTable(B)
var 集合相减 = function(A,B){
	for(n=1;#B;1){
		for(i=1;#B;1){
			if(table.find(A,B[i])!=null){
				table.removeByValue(A,B[i])
			}	
		
		}
	}
return A; 	
}
console.log("集合相减(A,B)：")
console.dumpTable(集合相减(A,B));
console.pause(true);


A = {5,3};
B = {1,2,3};
console.log("表A:")
console.dumpTable(A)
console.log("表B:")
console.dumpTable(B)
var 打包元组 = function(A,B){
	var C = {};
	for(k,v in A){
		C[v]=B[k]	
	}
	return C; 
}
console.log("打包元组(A,B)：")
console.varDump(打包元组(A,B));
console.pause(true);

Jackson

请教一下楼主，【集合相减】里，为何 for(n=1;#B;1)循环里再套一个同样的循环呢？