嗨这是我第一次使用这个网站，我做了一些关于如何将小写字母转换为大写字母但仍然是filles的研究。要求是检查“偶数”是否将偶数位字母转换为不同类型（从下到上或从上到下）。下面是我的代码：
function question4(str,pos)
{ var newLetter;
var kkk=str;
if (pos='even')
{
for (var i=0;i<str.length;i=i+2)
{
if (str[i].toString()==str[i].toString().toUpperCase())
{
newLetter=str[i].toString().toLowerCase();
kkk[i]=newLetter;
}else
{
newLetter=str[i].toUpperCase();
kkk[i]=newLetter;
}
}
}else if (pos='odd')
for ( i=0;i<str.length;i=i+2)
{
if (str[i]===str[i].toLowerCase())
{
alert('3');
}else if (str[i]===str[i].toUpperCase())
{
alert('4');
}
}
return kkk;
}
要求是：编写一个函数，根据匹配pos参数函数值的位置，改变字符串中所有字符的大小写。函数（str，pos [even | odd]）。示例（（'abCd'，'odd'）返回Abcd）
更新：现在我已经让“奇怪”状态正常工作，但“偶数”仍然无效，任何人都可以看看为什么？
function question4(strr,pos) {
var result ;
var sum="";
var aaa;
for (var i = 0; i <= strr.length - 1; i = i + 1)
{
if (pos == "odd"&&i%2==0)
{ aaa=strr.charCodeAt(i);
if (aaa >= 65 && aaa <= 90 )
{
result = String.fromCharCode(aaa + 32);
} else
result = String.fromCharCode(aaa - 32);
}
else if (pos == "even"&&i%2==1)
{
if (aaa >= 65 && aaa <= 90 )
{
result= String.fromCharCode(aaa + 32);
} else
result = String.fromCharCode(aaa - 32);
}else result=strr[i];
sum+=result;
}
return sum;
}
答案 0 :(得分：0)
为实现此目的，您可以通过char：char：＃/ p>来构造字符串
function question4(strInput, pos) {
let str = ""; // The string to construct
if (!pos || (pos !== "even" && pos !== "odd")) { // Validating pos
throw "invalid pos";
}
for (var i=0;i<strInput.length;i++) // Looping on strInput
{
let huPos = i + 1;
if ((pos === "even" && huPos%2 == 1) ||
(pos === "odd" && huPos%2 == 0)) {
/* If we want switch odd and we are on even position or if we want switch even and we are on odd position, then we add the original char
*/
str += strInput[i];
}
else {
// In others case, we switch lower to upper and upper to lower
let char = strInput[i];
str += char == char.toUpperCase() ? char.toLowerCase() : char.toUpperCase();
}
}
return str;
}
console.log(question4('abCdef', "odd")); // Return "AbcdEf"
编辑：
看到编辑后，我可以看到你想要在不使用toLower / UpperCase的情况下进行编辑。正如评论中所述，我认为这在js中是一个坏主意，但要进行实验，您可以实现这一点：
const reverser = {
"a": "a".charCodeAt(0),
"z": "z".charCodeAt(0),
"A": "A".charCodeAt(0),
"Z": "Z".charCodeAt(0),
};
const conversionValueToLower = reverser.a - reverser.A;
const conversionValueToUpper = reverser.A - reverser.a;
function reverseChar(char) {
var code = char.charCodeAt(0);
// If you want to go from upper to lower
if (code >= reverser.A && code <= reverser.Z) {
// Simply add the difference between lower and upper
return String.fromCharCode(code + conversionValueToLower);
} // Same logic here
else if (code >= reverser.a && code <= reverser.z) {
return String.fromCharCode(code + conversionValueToUpper);
}
/**
Or use if you want full digit
if (code <= 90 && code >= 65) {
return String.fromCharCode(code + 32);
}
else if (code >= 97 && code <= 122) {
return String.fromCharCode(code - 32);
}
**/
return char; // Other case return original char
}
function question4(strInput, pos) {
let str = "";
if (!pos || (pos !== "even" && pos !== "odd")) {
throw "invalid pos";
}
for (var i=0;i<strInput.length;i++)
{
let huPos = i + 1;
if ((pos === "even" && huPos%2 == 1) ||
(pos === "odd" && huPos%2 == 0)) {
str += strInput[i];
}
else {
str += reverseChar(strInput[i]);
}
}
return str;
}
console.log(question4('abCdef', "odd")); // return "AbcdEf"
另一种方法是编写模仿toLower / UpperCase
的utils函数我在你的答案中也纠正了你的代码，而没有改变原始逻辑
function question4(strr,pos) {
var result ;
var sum="";
var aaa;
for (var i = 0; i <= strr.length - 1; i = i + 1)
{
if (pos == "odd"&&i%2==0)
{ aaa=strr.charCodeAt(i);
if (aaa >= 65 && aaa <= 90 )
{
result = String.fromCharCode(aaa + 32);
} else if(aaa >=97&&aaa <=122)
{ result = String.fromCharCode(aaa - 32);}
else {result=strr[i];}
}
else if (pos == "even"&&i%2==1)
{ aaa=strr.charCodeAt(i);
if (aaa >= 65 && aaa <= 90 )
{
result= String.fromCharCode(aaa + 32);
} else if(aaa >=97&&aaa <=122)
{ result = String.fromCharCode(aaa - 32);}
else {result=strr[i];}
}else {result=strr[i];}
sum+=result;
}
return sum;
}
console.log(question4("abCd", "odd")) // return Abcd;
答案 1 :(得分：0)
这个问题的简单解决方案
// Function used to invert the letter case
const changeCase = c => {
if (c === c.toUpperCase()) return c.toLowerCase()
return c.toUpperCase()
}
const swapCaseConditional = (str, pos) => {
// Use split method to convert string into array and map the array
return str.split('').map((c, index) => {
if (pos === 'even') {
// if pos and index are even, change the letter case
if (index % 2) return changeCase(c)
return c
}
else {
// if pos and index are odd, change the letter case
if (!(index%2)) return changeCase(c)
return c
}
// Convert to string
}).join('')
}
console.log(swapCaseConditional('abCd', 'odd'))
答案 2 :(得分：0)
我工作了两个晚上，终于开始工作了。虽然没有完全涵盖所有情况，但几乎就在那里。
function question4(strr,pos) {
var result ;
var sum="";
var aaa;
for (var i = 0; i <= strr.length - 1; i = i + 1)
{
if (pos == "odd"&&i%2==0)
{ aaa=strr.charCodeAt(i);
if (aaa >= 65 && aaa <= 90 )
{
result = String.fromCharCode(aaa + 32);
} else
result = String.fromCharCode(aaa - 32);
}
else if (pos == "even"&&i%2==1)
{ aaa=strr.charCodeAt(i);
if (aaa >= 65 && aaa <= 90 )
{
result= String.fromCharCode(aaa + 32);
} else if(aaa >=97&&aaa <=122)
{ result = String.fromCharCode(aaa - 32);}
else {result=strr[i];}
}else {result=strr[i];}
sum+=result;
}
return sum;
}