密码要求:
1.长度超过8位
2.包括：大写字母/小写字母/数字/其它符号，以上四种至少三种
3.不能分割出两个相等的长度大于 2 的子串，例如 abcabc 可以分割出两个 abc，不合法，ababa 则无法分割出2个aba。
注：其他符号不含空格或换行
数据范围：输入的字符串长度满足 
[编程题]密码验证合格程序
const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
void (async function () {
while ((line = await readline())) {
// 长度不超过8
if (line.length <= 8) {
console.log("NG");
return;
}
// 不能有长度大于2的包含公共元素的子串重复
const length = line.length
const arr = []
for(let i = 0; i < length - 1; i++) {
// 将字符串切割成长度3的数组
arr.push(line.substring(i, i+3))
}
// 去重后对比是否有重复
const setArr = [...new Set(arr)]
if(setArr.length !== arr.length) {
console.log("NG");
return;
}
let num = false; // 数字
let big = false; // 大写字母
let small = false; // 小写字母
let other = false; // 其他符号
const bigReg = /[A-Z]/;
const smallReg = /[a-z]/;
for (let i = 0; i < line.length; i++) {
const isNum = Number.isNaN(Number(line[i]));
if (!isNum) {
num = true;
} else if (bigReg.test(line[i])) {
big = true;
} else if (smallReg.test(line[i])) {
small = true;
} else {
other = true;
}
}
let count = 0;
if (num) count += 1;
if (big) count += 1;
if (small) count += 1;
if (other) count += 1;
console.log(count >= 3 ? 'OK' : 'NG')
}
})();
发表于 2024-05-09 10:59:33
回复(0)
const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
void async function () {
/**
* 思路：
* 1. 长度超过8，判断长度
* 2. 包含四种符号中的三种，检测字符串有多少种符号
* 3. 寻找字符串中所有的子串，检查有没有相同的长度大于2的子串
* -> 搜索长度从3开始,用Set查重
* -> 子串长度i范围：3 ～ (len - 3)
*/
let caB = /[A-Z]{1,}/ // 大写字母
let lab = /[a-z]{1,}/
let dig = /\d{1,}/ // 数字
let other = /[^a-zA-Z0-9\s\n]{1,}/ // 其它字符
while(line = await readline()){
let markNum = 0
let reMark = 'OK'
if (line.length < 8) {
reMark = 'NG'
console.log(reMark) // 为了通过测例
continue
}
if (caB.test(line)) markNum++
if (lab.test(line)) markNum++
if (dig.test(line)) markNum++
if (other.test(line)) markNum++
if (markNum < 3) {
reMark = 'NG'
console.log(reMark)
continue
}
let sliceSet = new Set()
// 列举子串
for (let i = 3; i <= line.length - 3; i++) {
for (let j = 0; j + i <= line.length; j++) {
// 从头开始截子串，子串长度从3开始,不会超过len - 3
let conStr = line.slice(j, i+j) // slice(sIdx, eIdx)
if (sliceSet.has(conStr)) reMark = 'NG'
else sliceSet.add(conStr)
}
}
console.log(reMark)
}
}()
发表于 2023-12-13 15:42:31
回复(0)
const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
void async function () {
// Write your code here
while(true){
let str=await readline();
if(!str) break;
if(str.length<=8){
console.log("NG");
continue;
}
let map={};
for(let i in str){
let num=str[i].charCodeAt();
if(num>=97&&num<=122) map.low_case_letters=true;
else if(num>=65&&num<=90) map.up_case_letters=true;
else if(num>=48&&num<=57) map.nums=true;
else map.others=true;
}
if(Object.keys(map).length<3){
console.log("NG");
continue;
}
let flag=true;
for(let i=0;i<str.length-3;i++){
let child=str.slice(i,i+3);
let num1=str.indexOf(child),num2=str.lastIndexOf(child);
if(num1!=num2){
flag=false;
console.log("NG");
break;
}
}
if(!flag) continue;
console.log("OK");
}
}()
发表于 2022-11-11 20:49:04
回复(0)
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
let linelist = [];
rl.on('line', function (line) {
linelist.push(line);
});
rl.on('close', function (line) {
for(let i =0;i<linelist.length;i++){
let el = linelist[i];
if(el.length < 8){
console.log('NG');
continue;
}
//由于不会复杂的正则，只能一步一步来了
let list = [];
if(/[a-z]/.test(el)){
list.push(1);
}
if(/[A-Z]/.test(el)){
list.push(2);
}
if(/[0-9]/.test(el)){
list.push(3);
}
if(/[\W]/.test(el)){
list.push(4);
}
if(list.length >= 3){
if(fn(el)){
console.log('OK')
}else {
console.log('NG')
}
}else {
console.log('NG')
}
}
});
function fn(line){
for(let i=0;i<line.length-3;i++){
let str = line.substr(i,3);
let sty = line.split(str);
//不等于2表示被多次分割了，即有相同的子集
if(sty.length !== 2){
return false;
}
}
return true;
}
发表于 2022-04-22 17:31:41
回复(0)
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
const arr = [];
rl.on('line', function (line) {
arr.push(line);
if (line === "") {
rl.close();
}
});
rl.on("close", function(){
arr.forEach((item) => {
// Condition 1
if(item.length <= 8) {
console.log("NG");
return;
}
// Condition 2
const dataInd = Number(/\d+/.test(item));
const lowerCaseInd = Number(/[a-z]+/.test(item));
const upperCaseInd = Number(/[A-Z]+/.test(item));
const otherCharInd = Number(/[^A-Za-z0-9\s]+/.test(item));
const sum = dataInd + lowerCaseInd + upperCaseInd + otherCharInd;
if (sum < 3) {
console.log("NG");
return;
}
// Condition 3
for (let i = 0; i < item.length; i++) {
const subStr = item.slice(i, i + 3);
const nextIndex = item.indexOf(subStr, i + 3);
if (nextIndex !== -1) {
console.log("NG");
return;
}
}
console.log("OK");
return;
});
});
发表于 2022-01-22 00:20:59
回复(0)
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
rl.on('line', function (line) {
//长度不大于8或者有相同长度大于2的重复子串，则NG
if(line.length<=8||/(.{3,}).*\1/.test(line)){
console.log("NG");
return;
}else{
let count=0;//符合规则的计数器，小于3，则NG
if(/[0-9]/.test(line)){
count++;
}
if(/[a-z]/.test(line)){
count++;
}
if(/[A-Z]/.test(line)){
count++;
}
if(/\W/.test(line)){
count++;
}
if(count<3){
console.log("NG");
return;
}else{
// for(let i=0;i+2<line.length;i++){
// for(let j=0;j+2<line.length;j++){
// //i!=j时，判断每一个字符是否相等，相等则NG
// if(i!=j&&line.charAt(i)==line.charAt(j) && line.charAt(i+1)==line.charAt(j+1) && line.charAt(i+2)==line.charAt(j+2)){
// console.log("NG");
// return;
// }
// }
// }
//相同长度大于2的重复子串
// if(/(.{3,}).*\1/.test(line)){
// console.log("NG");
// return;
// }
console.log("OK");
return;
}
}
});
发表于 2021-11-25 23:51:48
回复(0)
let line
const getTypes = password => {
let count = 0;
/[0-9]/.test(password) && count++;
/[a-z]/.test(password) && count++;
/[A-Z]/.test(password) && count++;
/[^A-Za-z0-9]/.test(password) && count++;
return count;
}
const hasRepeat = password => {
let arr = password.split('')
for(let i = 0; i < password.length - 2; i++) {
let subStr = arr.splice(i, 3, ' ').join('')
if(arr.join('').includes(subStr)) return false
arr = password.split('')
}
return true
}
while(line = readline()) {
let length = line.length > 8
let type = getTypes(line) >= 3
let repeat = hasRepeat(line)
let ans = (length && type && repeat) ? 'OK' : 'NG'
print(ans)
}
编辑于 2021-07-08 11:54:24
回复(0)
while(line = readline()){
if (checkLength(line) &&checkCharkinds(line) && checkRepeat(line)) {
console.log('OK');
} else {
console.log('NG');
}
}
function checkLength(stringPass) {
if (stringPass == null || stringPass.length <= 8) {
return false;
} else {
return true;
}
}
// 2.包括大小写字母.数字.其它符号,以上四种至少三种
function checkCharkinds(stringPass) {
var digit = 0,
lowercase = 0,
uppercase = 0,
others = 0;
for (var i = 0; i < stringPass.length; i++) {
if (stringPass[i] >= '0' && stringPass[i] <= '9') {
digit = 1;
continue;
} else if (stringPass[i] >= 'a' && stringPass[i] <= 'z') {
lowercase = 1;
continue;
} else if (stringPass[i] >= 'A' && stringPass[i] <= 'Z') {
uppercase = 1;
continue;
} else {
others = 1;
continue;
}
}
var total = digit + lowercase + uppercase + others;
return total >= 3 ? true : false;
}
// 3.不能有相同长度超2的子串重复
function checkRepeat(stringPass) {
for (var i = 0; i < stringPass.length - 2; i++) {
var substr1 = stringPass.slice(i, i + 3);
if (stringPass.indexOf(substr1) != stringPass.lastIndexOf(substr1)) {
return false;
}
}
return true;
}
if (checkLength(line) &&checkCharkinds(line) && checkRepeat(line)) {
console.log('OK');
} else {
console.log('NG');
}
}
function checkLength(stringPass) {
if (stringPass == null || stringPass.length <= 8) {
return false;
} else {
return true;
}
}
// 2.包括大小写字母.数字.其它符号,以上四种至少三种
function checkCharkinds(stringPass) {
var digit = 0,
lowercase = 0,
uppercase = 0,
others = 0;
for (var i = 0; i < stringPass.length; i++) {
if (stringPass[i] >= '0' && stringPass[i] <= '9') {
digit = 1;
continue;
} else if (stringPass[i] >= 'a' && stringPass[i] <= 'z') {
lowercase = 1;
continue;
} else if (stringPass[i] >= 'A' && stringPass[i] <= 'Z') {
uppercase = 1;
continue;
} else {
others = 1;
continue;
}
}
var total = digit + lowercase + uppercase + others;
return total >= 3 ? true : false;
}
// 3.不能有相同长度超2的子串重复
function checkRepeat(stringPass) {
for (var i = 0; i < stringPass.length - 2; i++) {
var substr1 = stringPass.slice(i, i + 3);
if (stringPass.indexOf(substr1) != stringPass.lastIndexOf(substr1)) {
return false;
}
}
return true;
}
发表于 2021-01-09 22:15:05
回复(0)
let pwd='';
while(pwd = readline()){
let result='NG';
if(pwd.length>8){
let type={
a:0,
b:0,
c:0,
d:0
};
for(let ch of pwd){
let asc=ch.charCodeAt();
if('A'.charCodeAt()<=asc && asc<='Z'.charCodeAt()){
type.a++;
}else if('a'.charCodeAt()<=asc && asc<='z'.charCodeAt()){
type.b++;
}else if('0'.charCodeAt()<=asc && asc<='9'.charCodeAt()){
type.c++;
}else{
type.d++;
}
}
let count=0;
Object.keys(type).map(key=>{
if(type[key]>0){
count++;
}
})
if(count>=3){
let flag='OK';
for(let i=0;i<pwd.length-3;i++){
for(let j=i+3;j<pwd.length;j++){
let str=pwd.substring(i,j);
if(pwd.split(str).length>2){
flag="NG";
break;
}
}
if(flag==='NG'){
break;
}
}
result=flag;
console.log(result);
}else{
console.log(result);
}
}else{
console.log(result);
}
}
编辑于 2020-10-27 12:21:06
回复(0)
通过正则匹配得到结果
function validation(str) {
if(str.length <= 8) return 'NG'
if(str.match(/(.{3,})(?=.{3,}\1)/g)) {
return 'NG'
}
let count = 0
if(str.match(/\d+/g)) {
count++
}
if(str.match(/[a-z]+/g)) {
count++
}
if(str.match(/[A-Z]+/g)) {
count++
}
if(str.match(/[^a-zA-Z0-9]/g)) {
count++
}
if(!count) {
return 'NG'
}
return 'AC'
}
console.log(validation('aa67aaaAc7'),validation('aa67aaaA&^c7'),validation('abc67abcA&^c7'))
发表于 2018-08-11 14:08:06
回复(0)
//1.长度超过8位
function checkLength(stringPass){ if(stringPass==null||stringPass.length<=8) { return false; }else{ return true; }
}
// 2.包括大小写字母.数字.其它符号,以上四种至少三种
function checkCharkinds(stringPass){ var digit=0,lowercase=0,uppercase=0,others=0; for(var i=0;i<stringPass.length;i++){ if(stringPass[i]>='0'&&stringPass[i]<='9'){ digit=1; continue; } else if(stringPass[i]>='a'&&stringPass[i]<='z'){ lowercase=1; continue; } else if(stringPass[i]>='A'&&stringPass[i]<='Z'){ uppercase=1; continue; } else{ others=1; continue; } } var total=digit+lowercase+uppercase+others; return total>=3?true:false;
}
// 3.不能有相同长度超2的子串重复
function checkRepeat(stringPass){ for(var i=0;i<stringPass.length-2;i++){ var substr1=stringPass.slice(i,i+3); if(stringPass.indexOf(substr1)!=stringPass.lastIndexOf(substr1)){ return false; } } return true; }
var readline = require('readline').createInterface(process.stdin,process.stdout);
readline.on('line',function(stringPass){
if(checkLength(stringPass)&&checkCharkinds(stringPass)&&checkRepeat(stringPass)){ console.log('OK'); }else{ console.log('NG'); }
})
发表于 2018-06-05 22:57:23
回复(0)
const readline = require('readline');
const rl =readline.createInterface(process.stdin,process.stdout);
rl.on('line',function(s){
console.log(s.length>8&&/[^a-zA-Z0-9]/.test(s)+/[a-z]/.test(s)+/[A-Z]/.test(s)+/\d/.test(s)>2&&!/(...).*\1/.test(s)?"OK":"NG")
})//蛋疼的输入输出，**的js版本
发表于 2017-06-06 11:01:01
回复(0)
var readline = require("readline");
var rl = readline.createInterface(process.stdin, process.stdout);
rl.on("line", function(str){
if(str.length <= 8){
console.log('NG');
}else{
var re = /(.{3,}).*\1/;
var re1 = /[a-z]/;
var re2 = /[A-Z]/;
var re3 = /[1-9]/;
var re4 = /[\W|_]/;
var num = 0;
if(re.test(str)){
console.log('NG');
}else{
if(re1.test(str)){num++;}
if(re2.test(str)){num++;}
if(re3.test(str)){num++;}
if(re4.test(str)){num++;}
if(num >= 3){
console.log('OK');
}else{
console.log('NG');
}
}
}
rl.close();
})
发表于 2016-10-19 23:44:10
回复(0)
nodejs version 正则，长度 ，重复子串判断var readline = require('readline').createInterface(process.stdin,process.stdout);
readline.on('line',function(ans){
check(ans)?console.log("OK"):console.log("NG");
})
function check(ps){
const regcap = /[A-Z]+/;
const regchar = /[a-z]+/;
const regnum = /[0-9]+/;
const other = /[^A-Za-z0-9]+/;
regarr = [regcap,regchar,regnum,other];
var condition1 = ps.length - 8>0;
var condition2 = regarr.map(function(i){return
i.test(ps)}).reduce(function(i,j){ return i+j})>=3;
var condition3 = (function(str){
for(var i = 0;i<str.length-2;i++){
var substr = str.slice(i,i+3);
if(str.indexOf(substr)!=str.lastIndexOf(substr)){
return false;
}
}
return true;
})(ps);
return condition1&&condition2&&condition3;
}
发表于 2016-08-29 10:39:23
回复(1)
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