Fun JavaScript question: Help the postman sort addresses
一位邮递员将要去送邮件，他拥有一份客户地址的列表，如下所示：
"123 Main Street St. Louisville OH 43071,432 Main Long Road St. Louisville OH 43071,786 High Street Pollocksville NY 56432"。
上面的示例包含了三个地址，他们都用逗号分割了开来。
比如说，"123 Main Street St. Louisville OH 43071"，这个地址，它的格式实际上是这样组成的。
"门牌号 街道名 城镇名 邮政编号"，其中邮政编号如"OH 43071"，它是固定的8位，前两位是大写字母，接着是一个空格，最后是五位数字。
这位邮递员想对列表按照邮政编码进行分类。
那么任务来了，编写一个函数，传入两个参数。
travel(list , zipCode)
list代表地址列表，zipCode代表邮政编码，返回值是一个如下格式的字符串：
"邮政编码:街道名和城镇名,街道名和城镇名,.../门牌号,门牌号,..."
如果给定的邮政编码在列表中无法找到，那么就返回如下：
"邮政编码:/"
以上面的地址列表为例：
r = "123 Main Street St. Louisville OH 43071,432 Main Long Road St. Louisville OH 43071,786 High Street Pollocksville NY 56432" travel(r, "OH 43071") \\ "OH 43071:Main Street St. Louisville,Main Long Road St. Louisville/123,432" travel(r, "NY 56432") \\ "NY 56432:High Street Pollocksville/786" travel(r, "NY 5643") \\ "NY 5643:/"
这道题的思路很简单，考察的是对字符串进行操作，能够使用正则更好，不使用正则也不复杂。
下面是我的实现，尤其需要注意的是，假如邮政编码格式不对（不是8位），那么就无需再进行字符串操作，直接返回。
function travel(r, zipcode) {
var addresses = r.split(",");
var result = zipcode + ":";
var houseNumber = [];
var streets = [];
if(zipcode.length == 8){
for(var i=0;i<addresses.length;i++){
var current = addresses[i];
var pos = current.indexOf(zipcode);
if(pos >= 0){
var firstBlank = current.indexOf(" ");
houseNumber.push(current.slice(0,firstBlank));
streets.push(current.slice(firstBlank + 1,pos - 1));
}
}
}
return result + streets.join(",") + "/" + houseNumber.join(",");
}以上就是JavaScript趣题：帮邮递员分类地址的内容，更多相关内容请关注PHP中文网（www.php.cn）！
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