我正在尝试用字母表中的下一个字母替换字符串的所有字母。
例如:a - > b或i - >学家
我的程序忽略了检查字母数组字母的if语句。当我尝试运行代码时,它将所有字母替换为“A”,即字母数组中的最后一个元素。
虽然效果不好,但我发现这个算法没有任何错误。那么为什么程序会忽略if语句?
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");
答案 0 :(得分:2)
它无法正常工作的原因是因为ans
数组已被修改,而您仍在检查它。
在这个循环中:
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,则ans[i]
将被更新,但是在迭代的下一个循环中,它可能会再次出现,因为您正在检查更新ans[i]
变量。
正如@ xianshenglu建议的那样,一旦找到正确的匹配,你可以通过添加break
来解决这个问题。
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
}
}
有关替代方法,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char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('')
答案 1 :(得分:0)
break
执行时遗失了if
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"));