比较运算符不起作用（Java 脚本）

Question

我正在尝试用字母表中的下一个字母替换字符串中的所有字母。

例如：a --> b 或 i --> j.

我的程序忽略了根据字母表数组检查字母的 if 语句。当我尝试运行代码时，它将所有字母替换为 "A"，字母表数组中的最后一个元素。

虽然效率低下，但我找不到这个算法的任何错误。那么为什么程序会忽略 if 语句呢？

function LetterChanges(str){ 
var alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","a"];
str = str.toLowerCase();
var ans = str.split("");

for(i = 0; i < ans.length; i ++)//Running through for each letter of the input string
{
    for(a = 0; a < 26; a++)//Checking each letter against the alphabet array
    {
        if(alphabet[a] == ans[i])
        {
          ans[i] = alphabet[a+1];
        }
    }
}
return ans;
}

LetterChanges("Argument goes here");

Answer 1

当 if 执行时你输了 break

function LetterChanges(str){ 
var alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","a"];
str = str.toLowerCase();
var ans = str.split("");

for(i = 0; i < ans.length; i ++)//Running through for each letter of the input string
{
    for(a = 0; a < 26; a++)//Checking each letter against the alphabet array
    {
        if(alphabet[a] == ans[i])
        {
          ans[i] = alphabet[a+1];
           break;
        }
    }
}
return ans;
}

console.log(LetterChanges("Argument goes here"));

Answer 2

之所以不起作用，是因为 ans 数组已修改，而您仍在检查它。

在此循环中：

for(a = 0; a < 26; a++)//Checking each letter against the alphabet array
{
    if(alphabet[a] == ans[i])
    {
      ans[i] = alphabet[a+1];
    }
}

如果发现 if 语句为真，ans[i] 将被更新，但在迭代的下一个循环中，它可能再次为真，因为您正在检查更新后的 ans[i] 变量。

正如@xianshenglu 所建议的，您可以通过添加一个 break 来解决这个问题，以便在找到正确的匹配项后从内部循环中退出。

for(a = 0; a < 26; a++) {
  if(alphabet[a] == ans[i]) {
    ans[i] = alphabet[a+1]
    // escape from the inner loop once a match has been found
    break
  }
}

如需另一种方法，您可以执行以下操作：

var result = str.toLowerCase().split('').map(ch => {
  var pos = alphabet.indexOf(ch)
  return pos >= 0 ? alphabet[pos + 1] : ch
}).join('')

如果你想摆脱alphabet数组，你可以使用字符代码。例如：

var result = str.toLowerCase().split('').map(ch => {
  var code = ch.charCodeAt(0)
  if(code < 96 || code > 122){ return ch }
  return String.fromCharCode((code - 96) % 26 + 97)
}).join('')

比较运算符不起作用（Java 脚本）

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