Find the longest continuous number string in the string
Find the longest continuous number string in the string
Time limit: 3 seconds
Space limit: 32768K
Title description
Please find the longest continuous number string in the string, and return the length of this string; if there is a continuous number string with the same length, return the last continuous number string;
Note: the number string only needs to be composed of numbers No order is required. For example, the length of the number string "1234" is less than the number string "1359055". If there is no number,
an empty string ("") is returned instead of NULL! (Note: no need to consider negative)
input Description:
String Description Output:
continuous string of numbers in all figures & length of the longest string of
Example 1
Input
abcd12345ed125ss123058789
Output
123 058 789
. 9
Source code
/********************************************************************************************
在字符串中找出连续最长的数字串
时间限制：3秒
空间限制：32768K
题目描述
请一个在字符串中找出连续最长的数字串，并把这个串的长度返回；如果存在长度相同的连续数字串，返回最后一个连续数字串；
注意：数字串只需要是数字组成的就可以，并不要求顺序，比如数字串“1234”的长度就小于数字串“1359055”，如果没有数字，
则返回空字符串（“”）而不是NULL！（说明：不需要考虑负数）
输入描述 :
字符串输出描述:
连续数字串&在所有数字串最长的长度
示例1
输入
abcd12345ed125ss123058789
输出
123058789
9
********************************************************************************************/
#include <iostream>
#include <string>
using namespace std;
int main()
{
string srt = {
"abcd155555555555552345ed125ss123058789" };
int len_max = 0;
int len0 = 0;
int index = 0;
for (int i = 0; i < srt.size(); i++)
{
if (srt[i] >= '0'&& srt[i] <= '9')
{
len0++;
if (len0 > len_max)
{
index = i;
len_max = len0;
}
}
else
{
len0 = 0;
}
}
index = index- len_max+1;
for (int i = index; i < index+ len_max; i++)
{
printf("%c", srt[i]);
}
if (len_max == 0)
{
printf("数字字符串为空\n");
printf(" ");
}
printf("\n");
printf("len_max=%d", len_max);
printf("len_max=%d", len_max);
}