//弄成字符串组，字符串长度对比
import java.util.Scanner;
public class Main{
public static String characterToSpace(String string){
char[] arrays = string.toCharArray();
for(int i = 0;i < arrays.length;i++){
if(!Character.isDigit(arrays[i]))
arrays[i] = ' ';
}
String result = "";
for(Character cc : arrays){
result += cc.toString();
}
return result;
}
public static String longestDigitString(String string){
String strings[] = characterToSpace(string).split("\\s+");
int maxLength = 0;
String result = "";
for(String ss:strings){
if(ss.length() > maxLength){
maxLength = ss.length();
result = ss;
}
}
return result;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while(in.hasNext()){
String string = in.nextLine();
System.out.println(longestDigitString(string));
}
}
}

/*算法思想：用max表示经过的数字长度最大值，count表示数字计数器，当为字母时重置为0
*end表示数字尾部，每次满足数字时，对max进行判断，当max小于于count时，更新max和end
*/
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()){
String str = scanner.nextLine();
int max = 0，count=0,end=0;
for(int i=0;i
if(str.charAt(i)>='0' && str.charAt(i)<='9'){
count++;
if(max
max= count;
end = i;
}
}else{
count = 0;
}
}
System.out.println(str.substring(end-max+1,end+1));
}
}
}
编辑于 2017-03-15 20:29:12
回复(14)
26
"""
看你们写的好累啊
python这种题目四行解决问题
"""
import re
input_val = raw_input()
split_char = re.split(r'\D*', input_val)
print max(split_char, key=len)
编辑于 2017-04-01 08:22:28
回复(13)
7
虽然简单到令人发指 但是很经典的算法 #include 
#include 
#include 
#include 
using namespace std;
int main(){
string ans = "", str;
cin>>str;
int i = 0, max_len = 0, strlen = str.length();
while(i < strlen){
if(str[i] < '0' || str[i] > '9') i++;
else{
int len = 0;
string tmp = "";
while(i < strlen && str[i] <= '9' && str[i] >= '0'){
tmp = tmp + str[i];
i++; len++;
}
if(len > max_len) {max_len = len; ans = tmp;}//memcpy(&ans, &tmp, sizeof(tmp)); }
}
}
cout<
return 0;
}
/*
abcd12345ed125ss123456789
*/
编辑于 2018-09-17 22:10:18
回复(1)
8
华为的笔试考的这道题：
python解法如下： a = input()
maxLen, curLen, maxStr, curStr = 0, 0, "", ""
for i, v in enumerate(a):
if v.isnumeric():
curLen += 1
curStr += v
if curLen >= maxLen:
maxLen = curLen
maxStr = curStr
else:
curLen = 0
curStr = ""
print(maxStr)
编辑于 2017-09-07 13:59:50
回复(2)
7
#include 
#include 
int main(){
int i,k,key,sum,max,len[300];
char a[300];
scanf("%s",a);
for(int j=0;j<300;j++){//初始化长度数组
len[j]=0;
}
for(i=0;i
k=i;
sum=0;
while(a[i]>='0' && a[i]<='9'){
i++;
sum++;
}
len[k]=sum;//将每个位置出现的数组串长度，保存在len[]中，没出现的为 0
}
max=0;
for(i=0;i
if(max
max = len[i]; //保存最长的数字串长度
key = i;//记录最长数字串， 初始地址
}
}
for(i=0;i
printf("%c",a[key++]);
}
printf("\0");
printf("\n");
return 0;
}
发表于 2017-02-24 22:33:07
回复(0)
4
#include
#include 
usingnamespacestd;
int main()
{
// insert code here...
string str;
cin>>str;
string res;
int length=0;
int pos=0;
for(int i=0,j=0;j
{
if(str[j]'9')
{
++j;
i=j;
}
else
{
++j;
if((j-i)>length)
{
pos=i;
length=j-i;
}
}
}
res=str.substr(pos,length);
cout<
}
思路：双指针，遇到数字的时候，更新pos起始位置和子串的长度
发表于 2018-07-12 01:37:24
回复(2)
4
$str=trim(fgets(STDIN));
$len=strlen($str);
preg_match_all("/[0-9]+/",$str,$num);
$max=$num[0][0];
for($i=0;$i
{
if($num[0][$i]>$max) $max=$num[0][$i];
}
echo $max;
发表于 2017-09-16 16:34:39
回复(0)
6
//C++也可以很简洁
#include
#include
using namespace std;
int main()
{
string str,res,cur;
cin>>str;
for(int i=0;i<=str.length();i++){
if(str[i]>='0' && str[i]<='9')cur+=str[i];
else{
if(res.length()
cur="";
}
}
cout<
return 0;
}
编辑于 2018-07-20 23:58:57
回复(9)
3
Python n = input().strip()
ans,temp = [],[]
for i in n:
if 48 <= ord(i) <= 57:
temp.append(i)
else:
ans.append(temp)
temp = []
ans.append(temp)
ans = sorted(list(filter(None,ans)),key = len)
print(''.join(ans[-1]))
发表于 2018-07-08 22:32:24
回复(0)
3
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
String[] arr = s.split("[\\D]+");
String result="";
int max = 0;
for(int i=0;i
if(arr[i].length()>max){
max=arr[i].length();
result = arr[i];
}
}
System.out.println(result);
}
}
发表于 2017-09-08 15:28:44
回复(3)
5
编辑于 2017-02-17 23:54:35
回复(6)
2
let line = readline(); let arr = line.charAt(0);
let max = "";
for(let j = 1;j
if(arr.charCodeAt(arr.length-1)+1==line.charCodeAt(j)){
arr = arr+line.charAt(j);
}else{
arr = line.charAt(j);
}
if(max.length<=arr.length){
max = arr;
}
}
console.log(max);
编辑于 2018-10-09 17:29:16
回复(0)
1
public static void main(String[] args) {
ArrayList strlsit = new ArrayList<>();
String stri ="abcd12345ed125ss123456789";
StringBuffer str = new StringBuffer();
int size=0,max=0;
for(int i=0;i
if(stri.charAt(i)>='0'&&stri.charAt(i)<='9') {
str.append(stri.charAt(i));
size++;
}else {
str = new StringBuffer();
size=0;
}
}
max=size>max?size:max;
strlsit.add(str);
for(StringBuffer s:strlsit) {
if(s.length()==max) {
System.out.println(s);
}
}
发表于 2020-07-28 20:31:26
回复(0)
1
import java.util.Scanner;
public class Main {
//获取最长的连续数字串
public static String maxDigitalStr(String str){
char[] arrayChar = str.toCharArray();//将str转换成字符数组
int maxDigiStrLen = 0;//最长连续数字串的长度
String maxDigiStr = "";//最长连续数字串
int tempStrLen = 0;//临时数字串长度
StringBuffer tempStr = new StringBuffer();//临时数字串
for(int i = 0; i 
//过滤掉非数字的字符
if(arrayChar[i] >= '0' && arrayChar[i] <= '9'){
tempStr.append(arrayChar[i]);
tempStrLen++;
if(tempStrLen > maxDigiStrLen){
maxDigiStrLen = tempStrLen;//更新 最长连续数字串长度
maxDigiStr = tempStr.toString();//更新 最长连续数字串
}
}else{
tempStrLen = 0;//临时数字串长度 置0
tempStr.setLength(0);//临时数字串 清空
}
}
return maxDigiStr;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String tempStr = sc.next();
System.out.println(maxDigitalStr(tempStr));
}
}
编辑于 2020-07-22 22:55:48
回复(0)
1
直接新建一个一维数组，存放升序的累计个数，然后找最大，并往前输出最大长度的子字符串即为所求
while True:
try:
s=input()
res=[1 for i in range(len(s))]
for i in range(1,len(s)):
if ord(s[i])==ord(s[i-1])+1:
res[i]=res[i-1]+1
mmax=max(res)
ind=res.index(mmax)
print(s[ind+1-mmax:ind+1])
except:
break
发表于 2019-09-23 21:19:42
回复(0)
1
import re
input_val = input()
split_char = re.split(r'\D*', input_val)
lst = []
for i in split_char:
lst.append(len(i))
print(split_char[lst.index(max(lst))])
'''
max(lst)找出lst中最大的元素的值
lst.index(max(lst))找出“max(lst)找出lst中最大的元素的值”对应的索引
lst的大小和split_char是一致的，所以直接在split_char使用索引即可
'''
发表于 2019-08-04 13:52:28
回复(0)
1
#include 
using namespace std;
int main()
{
string a;
cin>>a;
int j=0;
int count=0;
int max=0;
int maxi;
for(int i=0;a[i]!='\0';i++)
{
if((int)a[i]<58&&(int)a[i]>47) //判断是不是0-9的数字
{
count++;
if(count>max)
{
max=count; //记录最大长度
maxi=i; //记录最大长度的结尾坐标
}
}
else //不满足条件，重新计数
{
count=0;
}
}
for(int i=maxi-max+1;i<=maxi;i++)
{
cout<
}
}
发表于 2019-04-21 20:26:11
回复(0)
1
/*
使用两个字符串，字符串sb记录当前数字串及长度，strTemp记录下一个数字串及长度，如
果strTemp长度比sb长，则替换sb，即sb=strTemp,最后输出sb
*/
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
StringBuilder sb = new StringBuilder("");
StringBuilder strTemp = new StringBuilder();
char[] ch = sc.nextLine().toCharArray();
for(int i=1;i
if(ch[i]>='0' && ch[i]<='9'){
strTemp.append(ch[i]);
}
if(ch[i]'9' || i == ch.length-1){
if(strTemp.length()>sb.length()){
sb = strTemp;
}
strTemp = new StringBuilder("");
}
}
System.out.println(sb);
}
}
}
发表于 2018-10-09 11:16:50
回复(0)
1
importjava.util.ArrayList;importjava.util.Scanner;public classSolution30 { public staticString MaxSubArray(char[] chars, intn) { if(n == 0) { return"";}
ArrayList curSum = newArrayList<>();ArrayList maxSum = newArrayList<>();StringBuilder sb = newStringBuilder();for(inti = 0;i < n;i++) { if(Character.isDigit(chars[i])) {
curSum.add(chars[i]);} else{ if(curSum.size() > maxSum.size()) {
maxSum.clear();for(intj = 0;j < curSum.size();j++) {
maxSum.add(curSum.get(j));}
}
curSum.clear();}
} for(inti = 0;i < maxSum.size();i++) {
sb.append(maxSum.get(i));} returnsb.toString();} public static voidmain(String[] args) {
Scanner in = newScanner(System.in);String str = in.nextLine();char[] chars = str.toCharArray();char[] newChars = new char[str.length() + 1];for(inti = 0;i < chars.length;i++) {
newChars[i] = chars[i];} intn = newChars.length;newChars[n - 1] = '#';System.out.println(MaxSubArray(newChars,n));}
}
发表于 2018-09-03 15:00:31
回复(0)
1
具体思想就是遍历去找数字，这个时间是标准O(n)的，关键在i=j这句， 遍历过的不用再遍历一次，因为比之前短的就不可能成为答案。往末尾加“a“是懒得边界判断了，加“a”很方便。 import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine().trim()+"a";
char []strs = line.toCharArray();
int tmpl = 0;
int tmpi = 0;
for(int i = 0; i< strs.length; i++){
if(Character.isDigit(strs[i])){
for(int j =i+1; j< strs.length; j++){
if (!Character.isDigit(strs[j])) {
if(j-i>tmpl){
tmpl = j-i;
tmpi = i;
}
i = j;
break;
}
}
}
}
System.out.println(line.substring(tmpi, tmpi+tmpl));
}
}