二进制转字符算法挑战
方法一:
function binaryAgent(str) {
var arr=str.split(" ");
for(var i in arr){
arr[i]=String.fromCharCode( parseInt(arr[i],2) );
}
var res=arr.join("");
return res;
}
binaryAgent("01000001 01110010 01100101 01101110 00100111 01110100 00100000 01100010 01101111 01101110 01100110 01101001 01110010 01100101 01110011 00100000 01100110 01110101 01101110 00100001 00111111");
方法二:
function binaryAgent(str) {
var arr = str.split(' '),
len = arr.length,
rStr = '';
for(var i=0;i<len;i++){
rStr += String.fromCharCode(parseInt(arr[i],2));
}
return rStr;
}
binaryAgent("01000001 01110010 01100101 01101110 00100111 01110100 00100000 01100010 01101111 01101110 01100110 01101001 01110010 01100101 01110011 00100000 01100110 01110101 01101110 00100001 00111111");