第 69 题： 如何把一个字符串的大小写取反（大写变小写小写变大写），例如 ’AbC' 变成 'aBc' 。

[zeroone001]

function processString (s) {
    var arr = s.split('');
    var new_arr = arr.map((item) => {
        return item === item.toUpperCase() ? item.toLowerCase() : item.toUpperCase();
    });
    return new_arr.join('');
}
console.log(processString('AbC'));

[iotale]

有没有想到用ascii码的？

const STEP = 'a'.charCodeAt() - 'A'.charCodeAt();

function transCase(char) {
  const asciiCode = char.charCodeAt();
  const resultCode = asciiCode < 'a'.charCodeAt() ? asciiCode + STEP : asciiCode - STEP;
  return String.fromCharCode(resultCode);
}

function transStr(str) {
  const charArr = str.split('');
  return charArr.map(char => transCase(char)).join('');
}

问题是只适用于大小写字母区间，不然表现会很怪异。

优化的时候可以做做边界条件。思路应该没问题。

[suguoyao]

马上想到的也是一楼的方法，另外也可以：

function swapString(str) {
  var result = ''

  for (var i = 0; i < str.length; i++) {
    var c = str[i]

    if (c === c.toUpperCase()) {
      result += c.toLowerCase()
    } else {
      result += c.toUpperCase()
    }
  }

  return result
}

swapString('ADasfads123!@$!@#') // => 'adASFADS123!@$!@#'

[suguoyao]

@iotale
你这个只考虑了字母，如果有其他字符就行不通了

transStr('adASjOdapASJO!@#4123123.l124')
// "ADasJoDAPasjoA`CTQRSQRSNLQRT"

[zhuzhh]

var str = 'AsdfsCsdefsd,$qwsDR'
var mapCode = {
    'A': 'A'.charCodeAt(0),
    'Z': 'Z'.charCodeAt(0),
    'a': 'a'.charCodeAt(0),
    'z': 'z'.charCodeAt(0),
}
var diff = mapCode.a - mapCode.A
var arr = str.split('')
for(var i = 0; i < arr.length; i++) {
    var code = arr[i].charCodeAt(0)
    if(code >= mapCode.A && code <= mapCode.Z) { // 大写
        arr[i] = String.fromCharCode(code + diff) // 转小写
    }else if(code >= mapCode.a && code <= mapCode.z){ // 小写
        arr[i] = String.fromCharCode(code - diff) // 转大写
    }else { // 不是字母
        console.log(arr[i], "不是字母")
        arr.splice(i, 1)
        i--
    }
}
console.log(arr.join(''))

[kungithub]

'AbcDefGh'.replace(/[a-zA-Z]/g,function(a){ return /[a-z]/.test(a)?a.toUpperCase():a.toLowerCase(); });

[YouziXR]

用ASCII做了一下。

// 第 69 题： 如何把一个字符串的大小写取反（大写变小写小写变大写），例如 ’AbC' 变成 'aBc' 。
const reverseStr = (str) => {
  let tmpAry = str.split('')
  let resultAry = []
  let a = 'a'.charCodeAt()
  let A = 'A'.charCodeAt()
  tmpAry.map((val) => {
    // debugger
    if (val <= 'Z' && val >= 'A') {
      resultAry.push(String.fromCharCode(val.charCodeAt() + (a - A)))
    } else if (val <= 'z' && val >= 'a') {
      resultAry.push(String.fromCharCode(val.charCodeAt() - (a - A)))
    } else {
      resultAry.push(val)
    }
  })
  return resultAry.join('')
}
console.log(reverseStr('aBCDefgh!@##%^$^*!@#$%'))

[btea]

[].map.call(str, function(item){     
     return /[a-z]/.test(item) ? item.toUpperCase() : item.toLowerCase(); 
}).join('');

[ZodiacSyndicate]

const reverse = str => {
        let result = ''
	for (let i = 0; i < str.length; i++) {
		if (str[i] > 'Z') {
			result += str[i].toUpperCase()
		} else {
			result += str[i].toLowerCase()
		}
	}
	return result
}

[Seasonley]

fn=(str)=>str.replace(/[A-Za-z]/g,c=>String.fromCharCode(c.charCodeAt()+(c<'['?32:-32)))
fn('-@A[Bc')//"-@a[bC"

[wangminglmm]

用了3种实现方法处理千万级别长度字符，对比了下时间，分享给大家，for循环+unicode转行相对较快
function getRandomStr() {
let str = ''
let arr = ["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", "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"
]
for (let i = 0; i < 10000000; i++) {
let num = parseInt(Math.random() * 53)
str += arr[num]
}
return str
}

    function transformStr1() {
        console.time('生成1千万长度字符串用时')
        let str = getRandomStr()
        console.timeEnd('生成1千万长度字符串用时')
        // console.log(str)
        console.time('使用正则替换用时')
        let str1 = str.replace(/./g, (input) => {
            let num = input.charCodeAt(0)
            if (num >= 65 && num <= 90) {
                return String.fromCharCode(num + 32)
            } else {
                return String.fromCharCode(num - 32)
            }
        })
        console.timeEnd('使用正则替换用时')
        // console.log(str)
        console.time('使用for循环耗时')
        let str2 = ''
        for (let i = 0, len = str.length; i < len; i++) {
            let s = str[i]
            let num = s.charCodeAt(0)
            if (num >= 65 && num <= 90) {
                str2 += String.fromCharCode(num + 32)
            } else {
                str2 += String.fromCharCode(num - 32)
            }
        }
        console.timeEnd('使用for循环耗时')
        console.log(str1 === str2)
        console.time('使用toUppercase逐个转换耗时')
        let str3 = ''
        for (let i = 0, len = str.length; i < len; i++) {
            let s = str[i]
            let sUp = s.toUpperCase()
            if (sUp === s) {
                str3 += s.toLocaleLowerCase()
            } else {
                str3 += sUp
            }
        }
        console.timeEnd('使用toUppercase逐个转换耗时')
        console.log(str2 === str3)
        console.time('使用toUppercase转换字符')
    }

[FounderIsShadowWalker]

let a = 'aB1cd';

const gap = 'a'.charCodeAt() - 'A'.charCodeAt();
const A_ASCII = 'A'.charCodeAt();
const Z_ASCII = 'Z'.charCodeAt();

const test = (a) => a.replace(/[a-zA-Z]/g, (str, value) => {
let ascii_code = str.charCodeAt();
if(ascii_code >= A_ASCII && ascii_code <= Z_ASCII){ //大写
return String.fromCharCode(ascii_code + gap) //转小写
}else{
return String.fromCharCode(ascii_code - gap);
}
})

console.error(test(a));

[xiatian]

正则实现：
str.replace(/(\w)/g, m => m === m.toUpperCase() ? m.toLowerCase() : m.toUpperCase())

[fuqianxi]

const str = 'AjklAJIBNiuh';
console.log(str)

let newStr = ''
for (const i of str) {
  let j = '';
  if (/[A-Z]/.test(i)) {
   j =  i.toLowerCase();
   
  } else {
   j = i.toUpperCase();
   
  }
  newStr+=j
}
console.log(newStr)

[Bottle1206]

不用转成数组直接replace操作字符串

function transfer(str) {
  return str.replace(/[a-zA-Z]/g, match => {
    return /[a-z]/.test(match) ? match.toUpperCase() : match.toLowerCase()
  })
}

[lxinr]

function transString(str) {
  if(typeof str !== 'string') return
  let strArr = str.split('')
  return strArr.reduce((acc, cur) => acc += (cur === cur.toUpperCase() ? cur.toLowerCase() : cur.toUpperCase()))
}