题解 | #字符串加解密 node环境解法#

const readline = require('readline')
const rl = readline.createInterface({
  input: process.stdin,
  output: process.stdout
})

function test(s1,s2){
  let res1 = ''
  let res2 = ''
  for(let i=0;i<s1.length;i++){
    if(Number(s1[i]) || s1[i] === '0'){
      if(Number(s1[i]) === 9){
        res1 += '0'
      }else{
        let temp = Number(s1[i])
        res1 += temp + 1
      }
    }else{
      // 首先判断字母是大写还是小写，因为要转换大小写
      if(/[a-z]/.test(s1[i])){ // 如果是小写字符走这条逻辑
        let n = s1[i].toUpperCase().charCodeAt()
        if(n === 90){
          res1 += 'A'
        }else{
          res1 += String.fromCharCode(n+1)
        }
      }else{ // 大写字符走这条逻辑
        let n = s1[i].toLowerCase().charCodeAt()
        if(n === 122){
          res1 += 'a'
        }else{
          res1 += String.fromCharCode(n+1)
        }
      }
    }
  }
  for(let i=0;i<s2.length;i++){
    if(Number(s2[i]) || s2[i] === '0'){
      if(Number(s2[i]) === 0){
        res2 += '9'
      }else{
        let temp = Number(s2[i])
        res2 += temp - 1
      }
    }else{
      // 首先判断该字符是大写还是小写
      if(/[a-z]/.test(s2[i])){ // 如果小写走这条逻辑
        let n = s2[i].toUpperCase().charCodeAt()
        if(n === 65){
          res2 += 'Z'
        }else{
          res2 += String.fromCharCode(n - 1)
        }
      }else{ // 大写走这条逻辑
        let n = s2[i].toLowerCase().charCodeAt()
        if(n === 97){
          res2 += 'z'
        }else{
          res2 += String.fromCharCode(n - 1)
        }
      }

    }
  }
  console.log(res1)
  console.log(res2)
}

const arr = []
rl.on('line',function(line){
  arr.push(line)
})

rl.on('close',function(){
  test(arr[0],arr[1])
  // 可以使用字符串的charCodeAt来判断字符的ASCII码，这道题主要用到的方法就是charCodeAt,String.fromCharCode
  // console.log(String.fromCharCode('A'.charCodeAt()))
  // console.log('A'.charCodeAt())
  // console.log('Z'.charCodeAt())
  // console.log('a'.charCodeAt())
  // console.log('z'.charCodeAt())
})