31 条回复
前端第一道（最长非公共字符串）：
package shunfeng920;
import java.util.Scanner;
/**
* Created by JackHui on 2017/9/20.
*/
public class no2 {
public static void main(String[] args) {
Scanner sc = new Scanner( System.in );
String str1=sc.nextLine();
String str2=sc.nextLine();
if(str2.contains( str1 )||str1.contains( str2 ))
{
System.out.println("0");
}
else if(str1.length()>str2.length())
{
System.out.println(str1.length());
}
else if(str2.length()>str1.length())
{
System.out.println(str2.length());
}
else if(str1!=str2)
{
System.out.println(str2.length());
}
else {
for(int i=0;i<str1.length();i++)
{
if(str2.indexOf( str1.charAt( i ) )<0)
{
System.out.println(str1.length());
break;
}
}
}
}
}第二道（密码）：
package shunfeng920;
import java.util.Scanner;
/**
* Created by JackHui on 2017/9/20.
*/
public class no1 {
public static void main(String[] args)
{
Scanner sc=new Scanner( System.in );
int x=sc.nextInt();
int[] num=new int[x];
for(int i=0;i<x;i++)
{
num[i]=sc.nextInt();
}
int[] num2=new int[x];
for(int j=0;j<x-1;j++)
{
num2[j]=num[j]+num[j+1];
System.out.println(num2[j]);
}
num2[x-1]=num[x-1];
System.out.println(num2[x-1]);
}
}
题目一：最长非公共子串
function solve(s1, s2) {
var len1 = s1.length;
var len2 = s2.length;
var max = (len1>=len2)?s1:s2;
var min = (len1>=len2)?s2:s1;
if(max.indexOf(min)===-1){return min.length;}
else if(max.indexOf(min)!==-1){
var index = max.indexOf(min);
var pre = max.substring(0,index).length;
var next = max.substring(index+min.length,max.length).length;
return pre>next?pre:next;
}
}题目二：木木的密码
function solve(a) {
var arr = new Array(a.length);
for(var i =0;i<a.length;i++){
arr[i] = a[i] + (a[i+1] || 0);
}
return arr;
}第二题幸运数字问题 没ac 只有30%
package shunfeng;
import java.util.Scanner;
/**
* Created by Administrator on 2017/9/20 0020.
*/
public class T2 {
static String IntToStr(int n){
StringBuilder stringBuilder=new StringBuilder();
if(n==0)return "0";
while(n>0){
stringBuilder.insert(0,n%2);
n=n/2;
}
return stringBuilder.toString();
}
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
while(in.hasNext()){
//求4和7组成的数字;小于N
Long n=in.nextLong();
// System.out.println(n);
long x=0;
int count=-1;
// 4 看成 0 7看成1即可
boolean flag=false;
int count1=-1;
while(x<=n){
count++;
count1++;
String s=IntToStr(count);
StringBuilder tem=new StringBuilder();
for (int i = 0; i <s.length() ; i++) {
if(s.charAt(i)=='0')tem.append(4);
else tem.append(7);
}
x=Long.parseLong(tem.toString());
System.out.println(x);
long z=0;
while(z<=n) {
s="0"+s;
tem=new StringBuilder();
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '0') tem.append(4);
else tem.append(7);
}
z=Long.parseLong(tem.toString());
if ( z<= n) {
count1++;
System.out.println("-----" + Integer.parseInt(tem.toString()));
}
}
}
System.out.println(count1);
}
}
}
幸运数
public static void main(String args[]){
Scanner reader = new Scanner(System.in);
int sum = 0;
int num = reader.nextInt();
for(int i = 0; i <= num; i ++){
int b = (i+"").length();
int a = 0;
char[] ch = (i+"").toCharArray();
for(int j = 0; j < ch.length; j++){
if((ch[j]+"").matches("[47]")){
a++;
}
}
if(a==b){
sum++;
}
}
System.out.println(sum);
}第二题幸运数问题
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Main {
/******************************开始写代码******************************/
static void solve(long num) {
long index = 0;
long n = 10;
long len = num/10;
long count[] = new long[(int) (len+1)];
while(num/n>0){
index++;
n = n*10;
}
long total = 1;
/***************************/
for(int i = 0;i<index;i++){
total = total *2;
count[i] = total;
}
//
total = 0;
for(int i = 0;i<index;i++){
total = total+count[i];
}
//
if(num<4){
System.out.println(0);
return;
}
if(num<7){
System.out.println(1);
return;
}
if(num<44){
System.out.println(2);
return;
}
while(index>=0){
n = n/10;
if(num/n<4){
System.out.println(total);
return;
}else if(num/n==4){
index--;
}else if(num/n<7){
System.out.println(total+count[(int) index-1]);
return;
}else if(num/n ==7){
total =total+count[(int)index-1];
index--;
}else if(num/n>7){
System.out.println(total+count[(int) index-1]*2);
return;
}
num = num-(num/n)*n;
if(num<10){
if(num<4){
System.out.println(total);
return;
}
if(num<7){
System.out.println(total+1);
return;
}
System.out.println(total+2);
return;
}
}
}
/******************************结束写代码******************************/
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
long num = scanner.nextLong();
scanner.close();
solve(num);
}
}
public class Main {
static int find2(String a, int index, int length) {
// 444->1 447->2 474->3 477->4 744->5 747->6 774->7 777->8
// 规律是末位的4贡献1，7贡献2，非末位的4无贡献，7贡献2^n，n为7后面的位数
char c = a.charAt(index);
if (index == length-1) {
if (c == '4') return 1;
else return 2;
}
if (c == '4') return find2(a, index+1, length);
else return (int)Math.pow(2, length-index-1) + find2(a, index+1, length);
}
static int find(String a) {
// 找到小于该整数的最大数，且该最大数只由4和7组成
if (a.charAt(0) < '4') return 0;
StringBuffer b = new StringBuffer(a.length());
char c;
boolean overflow = false;
for (int i = 0; i < a.length(); i++) {
if (overflow) {
b.append("7");
continue;
}
c = a.charAt(i);
if (c < '4') {
while (i > 0 && b.charAt(--i) != '7') {
b.deleteCharAt(i);
}
if (b.length() == 0) return 0;
b.setCharAt(i, '4');
overflow = true;
}
else if (c == '4') b.append("4");
else if (c < '7') {
b.append("4");
overflow = true;
}
else if (c == '7') b.append("7");
else {
b.append("7");
overflow = true;
}
}
return find2(b.toString(), 0, b.length());
}
static int solve(String a) {
int ret = 0;
int size = a.length();
int MOD = 1000000007;
int up = find(a); // 长度与a相同的幸运数
// 长度比a小的幸运数
for (int i = 1; i < size; i++) {
ret += Math.pow(2, size-i);
ret %= MOD;
}
return (ret + up) % MOD;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (true) {
String bigNum = in.nextLine();
int ret = solve(bigNum);
System.out.println(ret);
}
}
}抛砖引玉一下，答完了才写出来的，也不知道对不对
package myPackage;
//顺丰科技测试题
public class Main {
public static void main(String[] args) {
for(int i = 0; i < 100; i++){
int limit = (int) (Math.random() * 2100000000); //21亿
System.out.println("数的上界: " + limit);
String numStr = String.valueOf(limit);
char[] numArray = numStr.toCharArray();
long start = System.currentTimeMillis();
int c1 = firstNum(numArray);
long end = System.currentTimeMillis();
long t1 = (end - start) > 0 ? (end - start) : 1;
System.out.println("算法的结果:" + c1 + ", " + "用时" + t1 + "ms");
start = System.currentTimeMillis();
int c2 = countLuckyNum(limit);
end = System.currentTimeMillis();
long t2 = (end - start) > 0 ? (end - start) : 1;
System.out.println("蛮力检验的结果:" + c2 + ", " + "用时" + t2 + "ms, 比算法慢了" +
String.format("%.2f", (1 - 1.0 / t2 / (1.0 / t1)) * 100) + "%" );
if(c1 != c2){
System.out.println("出现错误!!!!!!!!!!!!!!!!!!!!!!!!");
break;
}
System.out.println();
}
System.out.println("算法正确");
}
//处理第一个数字的取值(因为如果第一个数字可取0,并且会导致紧跟着的后面的数字可取0, 所以要区分对待第一个数字和其他数字)
static int firstNum(char[] arr){
int n = whichNum(arr[0]);
int len = arr.length;
if(len == 1){
if(n > 7 || n == 7)
return 2;
else if((n > 4 || n == 4) && n < 7)
return 1;
else
return 0;
}
if(n > 7)
return (int) (Math.pow(2, len + 1) - 2);
else if(n == 7){
int testValue = nextNum(arr, 1);
int certainValue = (int) (3 * Math.pow(2, len - 1) - 2);
if(testValue == -1)
return certainValue;
else
return testValue + certainValue;
}
else if(n > 4 && n < 7){
return (int) (3 * Math.pow(2, len - 1) - 2);
}
else if(n == 4){
int testValue = nextNum(arr, 1);
int certainValue = (int) (Math.pow(2, len) - 2);
if(testValue == -1)
return certainValue;
else
return testValue + certainValue;
}
else
return (int) (Math.pow(2, len) - 2);
}
//处理第一个数字后面的数字
static int nextNum(char[] arr, int p){
if(p == arr.length)
return 1;
else{
int r = arr.length - p;
int n = whichNum(arr[p]);
if(n > 7)
return (int) Math.pow(2, r);
else if(n == 7){
int testValue = nextNum(arr, p + 1);
int certainValue = (int) (Math.pow(2, r - 1));
if(testValue == -1)
return certainValue;
else
return testValue + certainValue;
}
else if(4 < n && n < 7){
return (int) Math.pow(2, r - 1);
}
else if(n == 4){
return nextNum(arr, p + 1);
}
else
return -1;
}
}
//数字字符转为数字
static int whichNum(char c){
switch(c){
case '1' : return 1;
case '2' : return 2;
case '3' : return 3;
case '4' : return 4;
case '5' : return 5;
case '6' : return 6;
case '7' : return 7;
case '8' : return 8;
case '9' : return 9;
default : return 0;
}
}
//下面用蛮力法验证算法的正确性
static int countLuckyNum(int limit){
int count = 0;
for(int i = limit; i > 3; i--){
if(isLuckyNum(i)){
count++;
}
}
return count;
}
static boolean isLuckyNum(int num){
int value;
while(num > 0){
value = num % 10;
if(value != 4 && value != 7)
return false;
num /= 10;
}
return true;
}
} static int topDigit(char[] arr){ //讨论最高位的值n
int n = arr[0] - '0';
int r = arr.length;
//此处不用专门讨论数字只有一位的情况, 因为也符合规律
int result = (int) (Math.pow(2, r) - 2); //countLessDigit
if(n > 7)
result += (int) (Math.pow(2, r));
else if(n == 7)
result += nextDigit(arr, 1) + (int)Math.pow(2, r - 1);
else if(n > 4)
result += (int)Math.pow(2, r - 1);
else if(n == 4)
result += nextDigit(arr, 1);
return result;
}
static int nextDigit(char[] arr, int pos){ //讨论最高位后面的较低位的值n
if(pos == arr.length)
return 1;
int r = arr.length - pos;
int n = arr[pos] - '0';
if(n > 7)
return (int) Math.pow(2, r);
else if (n == 7)
return (int)Math.pow(2, r - 1) + nextDigit(arr, pos + 1);
else if (n > 4)
return (int) Math.pow(2, r - 1);
else if(n == 4)
return nextDigit(arr, pos + 1);
else
return 0;
}
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