Question

    var str = "ajdisoiureenvmcnmvm"
    var arr1 = ["e","t","a","i","n","o","h","r","d","q","l","c","u","m","w","f","s","g","y","p","b","v","k","j","x","z"]
    var arr2 = ["m","i","e","n","v","r","d","j","s","o","u","c","b","a","f","g","h","k","l","p","q","t","w","x","z","y"]

更换字符串的最有效方法是什么，如果它包含“m”，则将其替换为“e”，将“i”替换为“t”，依此类推。

Answer 1

如果您对map：

没问题

function encode (string) {
  return string.split("").map(function (letter) {
    return arr1[arr2.indexOf(letter)];
  }).join("");
}

Answer 2

这取决于您对“有效”的定义。有了这种结构，就很难有效率。但是像这样的东西是功能和简短：

var rex = new RegExp("[" + arr1.join("") + "]", "g");
var result = str.replace(rex, function(letter) {
    var index = arr1.indexOf(letter);
    return index === -1 ? letter : arr2[index];
});

Live Example | Live Source

在那里，我们使用字符类创建一个正则表达式来匹配每个源字符，然后使用函数执行替换操作以从数组中查找替换（如果有）。

现在，如果你可以改变结构，有更好的方法，比如使用地图而不是arr1.indexOf。 E.g：

var map = {
    "e": "m",
    "t": "i",
    "a": "e",
    // ...
};

然后（这使用ES5的Object.keys，如果需要可以填充）：

var rex = new RegExp("[" + Object.keys(map).join("") + "]", "g");
var result = str.replace(rex, function(letter) {
    return map[letter] || letter;
});

Answer 3

这应该是自我解释的并且相对有效。

str = str.split("")
for(var i = 0; i < str.length; i++) {
    var c = str[i];
    var j = arr1.indexOf(c);
    str[i] = arr2[j];
}
str = str.join("");

我会使用映射来处理这种事情。

Answer 4

var str = "ajdisoiureenvmcnmvm"
var arr1 = ["e","t","a","i","n","o","h","r","d","q","l","c","u","m","w","f","s","g","y","p","b","v","k","j","x","z"]
var arr2 = ["m","i","e","n","v","r","d","j","s","o","u","c","b","a","f","g","h","k","l","p","q","t","w","x","z","y"]

newStr = str.split("");
for (var i=0, len=newStr.length; i<len; i++)
{
    index = arr2.indexOf(newStr[i]);
    if(index>=0)
        newStr[i] = arr1[index];
}
str = newStr.join("");
console.log(str);

Answer 5

可能更直接的方法是先将原始字符串放入数组中，这样您就可以更加轻松地逐个操作它：

var arr1 = ["e","t","a","i","n","o","h","r","d","q","l","c","u","m","w","f","s","g","y","p","b","v","k","j","x","z"];
var arr2 = ["m","i","e","n","v","r","d","j","s","o","u","c","b","a","f","g","h","k","l","p","q","t","w","x","z","y"];
var str = "ajdisoiureenvmcnmvm";
var data = str.split(""), index;
for (var i = 0; i < arr2.length; i++) {
    index = str.indexOf(arr2[i]);
    if (index >= 0) {
        data[index] = arr1[i];
    }
}
// build the string back again
str = data.join("");

工作演示：http://jsfiddle.net/jfriend00/fAQTe/

Answer 6

我知道你没有要求shell脚本的方法，但它会让你觉得在这个特殊请求中tr命令可以轻松完成。

echo "ajdisoiureenvmcnmvm" |tr "etainohrdqlcumwfsgypbvkjxz" "mienvrdjsoucbafghklpqtwxzy"

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用两个数组替换字符串中的字母？

6 个答案: