We have given a string str containing the numeric and alphabetical characters. We need to find the sum of all numbers represented by a continuous sequence of digits available in the given string.
str = “12were43”
55
The sum of 12 and 43 is equal to 55.
str = “1a2c3d”
6
1、2和3的和为6。
str = “werderfrewsf”
0
It gives 0 in the output as the string contains no digit.
我们解决问题的逻辑是从给定的字符串中提取所有数字并求和。
In this approach, we will use isDigit() method to check whether the current character is a digit. Also, we multiply the current value of the number by 10 and add the current character to the number if the current character is a digit.
步骤 1 - 将 'number' 和 'sum' 变量初始化为零。
Step 2 − Iterate through the string and check current character is between 0-9 using the isDigit() method.
步骤 3 - 如果当前字符是数字，则将数字值乘以10，并加上当前数字值。
第四步 - 如果当前字符不是数字，则将“number”变量的值添加到“sum”变量中，并将“number”变量的值更新为零。
Step 5 − Once the iteration of the loop completes, add the value of the ‘number’ to the ‘sum’ variable and return the value of the sum variable.
#include <bits/stdc++.h>
using namespace std;
// function to return the sum of the consecutive number present in the string
int getSumOfDigits(string str){
// store the current number
int number = 0;
// Stores total sum
int sum = 0;
// Traverse the string
for (auto &ch : str){
// If the current character is between '0' and '9', append it to the number
if (isdigit(ch)) {
number = number * 10 + ch - '0';
} else {
// if the current character is not between '0' and '9', add 'number' to the sum and reset 'number'
sum += number;
number = 0;
}
}
// if the number is greater than 0, add it to sum
sum += number;
return sum;
}
int main(){
string str = "6we24er5rd6";
cout << "The sum of consecutive digits in the given string is - " << getSumOfDigits(str);
return 0;
}
The sum of consecutive digits in the given string is - 41
时间复杂度 - O(n)，因为我们只使用了一个循环。
空间复杂度 − O(1)，因为我们不使用任何额外的空间。
In this approach, we use the ASCII values of the character to check whether the current character is a digit. Also, we append characters to the ‘number’ variable until we get digits in the string and use the atoi() method to extract the number from the string.
步骤1 - 定义'number'变量并将其初始化为空字符串。同时，定义'sum'变量并将其初始化为0。
Step 2 − Use for loop to traverse the string and get each character of the string.
步骤 3 - 如果 c-‘0’ 大于等于零且小于等于 9，则表示当前字符是一个数字。
Step 4 − If the current character is a digit, append it to the ‘number’ string.
Step 5 − If the current character is not a digit, use the c_str() method to convert the number string to a character array and pass it as a parameter of the atoi() method to convert the string to a number. Also, update the number string with the “” value.
The atoi() method returns a number if the string is convertible to a number; Otherwise, it returns zero.
Step 6 − Once the iteration of for loop completes, again use the atoi() method to convert the string to a number and add to the sum value.
#include <bits/stdc++.h>
using namespace std;
// function to return the sum of the consecutive numbers present in the string
int getSumOfDigits(string str){
string number = "";
// to store the sum of all the consecutive numbers
int sum = 0;
// traverse the string
for (char c : str){
// if the current character is between 0 to 9
if (c - '0' >= 0 && c - '0' <= 9){
// append it to the number string
number += c;
}
// if the current character is an alphabet
else {
// convert string to an array of characters and pass it to atoi() function
sum += atoi(number.c_str());
// reset temporary string to empty
number = "";
}
}
// if the number is greater than 0, add it to sum
sum += atoi(number.c_str());
return sum;
}
int main(){
string str = "11aa32bbb5";
cout << "The sum of consecutive digits in the given string is - " << getSumOfDigits(str);
return 0;
}
The sum of consecutive digits in the given string is - 48
时间复杂度 - O(N)
空间复杂度 − O(1)
在这种方法中，我们使用正则表达式来找到所有数字的匹配项。之后，我们可以将字符串转换为数字并将其添加到sum变量中。
Step 1 − Define the regex pattern.
Step 2 − Use the regex_search() method to find the match for the number string.
Step 3 − Make iterations using a while loop as long as we find matches.
Step 4 − In the while loop, use the stoi() method to convert the string to a number and add it to the sum variable.
第5步 - 同样，使用match().suffix()方法更新字符串。这样我们就不会得到重复的匹配。
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the sum of the numbers found in the string
int getSumOfDigits(string str){
// regex pattern to find the numbers in the string
regex pattern("d+");
smatch match;
// variable to store the sum of the numbers
int sum = 0;
// using the regex_search() function to find the numbers
while (regex_search(str, match, pattern)){
// adding the numbers to the sum variable
sum += stoi(match[0].str());
// update the string
str = match.suffix().str();
}
return sum;
}
int main(){
// input alphanumeric string
string str = "abc23@12";
cout << "The sum of consecutive digits in the given string is - " << getSumOfDigits(str);
return 0;
}
The sum of consecutive digits in the given string is - 0
Time complexity − O(N), as regex finds matches by iterating through the string.
空间复杂度 − O(1)
我们学习了三种不同的方法来找到字符串中连续数字的和。最后一种方法是最优化的代码，因为它使用了正则表达式。然而，对于初学者来说，使用正则表达式可能会很困难。
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