求字符串字母值的一位数和
给定大小为 N 的字符串 S ，任务是通过给定字符串中所有字母的顺序之和得到的值的重复位数来求一位数之和。
字母的顺序是由它们在英语中出现的位置决定的。
示例:
输入: S =【极客】 输出: 1 解释: 字母的和顺序得到的值是 7 + 5 + 5 + 11 = 28。 28 = 2+8 = 10 = 1+0 = 1 相加得到的一位数和。
输入:S = " geeks forgeeks " T3】输出: 7
方法:给定的问题可以这样解决:如果 S[i] 是小写或大写字符，则通过将值(S[I]–‘A’+1)或(S[I]–‘A’+1)相加，首先找到给定字符串中存在的所有字母的顺序的和。找到和的值后，使用本文中讨论的方法找到该值的个位数。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
int findTheSum(string str)
{
string alpha;
// Traverse the given string
for (int i = 0; i < str.length(); i++) {
// If character is an alphabet
if ((str[i] >= 'A' && str[i] <= 'Z')
|| (str[i] >= 'a' && str[i] <= 'z'))
alpha.push_back(str[i]);
}
// Stores the sum of order of values
int score = 0, n = 0;
for (int i = 0; i < alpha.length(); i++) {
// Find the score
if (alpha[i] >= 'A' && alpha[i] <= 'Z')
score += alpha[i] - 'A' + 1;
else
score += alpha[i] - 'a' + 1;
}
// Find the single digit sum
while (score > 0 || n > 9) {
if (score == 0) {
score = n;
n = 0;
}
n += score % 10;
score /= 10;
}
// Return the resultant sum
return n;
}
// Driver Code
int main()
{
string S = "GeeksforGeeks";
cout << findTheSum(S);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
static int findTheSum(char []str)
{
String alpha="";
// Traverse the given String
for (int i = 0; i < str.length; i++) {
// If character is an alphabet
if ((str[i] >= 'A' && str[i] <= 'Z')
|| (str[i] >= 'a' && str[i] <= 'z'))
alpha+=(str[i]);
}
// Stores the sum of order of values
int score = 0, n = 0;
for (int i = 0; i < alpha.length(); i++) {
// Find the score
if (alpha.charAt(i) >= 'A' && alpha.charAt(i) <= 'Z')
score += alpha.charAt(i) - 'A' + 1;
else
score += alpha.charAt(i) - 'a' + 1;
}
// Find the single digit sum
while (score > 0 || n > 9) {
if (score == 0) {
score = n;
n = 0;
}
n += score % 10;
score /= 10;
}
// Return the resultant sum
return n;
}
// Driver Code
public static void main(String[] args)
{
String S = "GeeksforGeeks";
System.out.print(findTheSum(S.toCharArray()));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python program for the above approach
def findTheSum(str):
alpha = ""
# Traverse the given string
for i in range(0, len(str)):
# If character is an alphabet
if ((str[i] >= 'A' and str[i] <= 'Z') or (str[i] >= 'a' and str[i] <= 'z')):
alpha += str[i]
# Stores the sum of order of values
score = 0
n = 0
for i in range(0, len(alpha)):
# Find the score
if (alpha[i] >= 'A' and alpha[i] <= 'Z'):
score += ord(alpha[i]) - ord('A') + 1
else:
score += ord(alpha[i]) - ord('a') + 1
# Find the single digit sum
while (score > 0 or n > 9):
if (score == 0):
score = n
n = 0
n += score % 10
score = score // 10
# Return the resultant sum
return n
# Driver Code
if __name__ == "__main__":
S = "GeeksforGeeks"
print(findTheSum(S))
# This code is contributed by rakeshsahni
C
// C# program for the above approach
using System;
class GFG {
static int findTheSum(string str)
{
string alpha = "";
// Traverse the given string
for (int i = 0; i < str.Length; i++) {
// If character is an alphabet
if ((str[i] >= 'A' && str[i] <= 'Z')
|| (str[i] >= 'a' && str[i] <= 'z'))
alpha += (str[i]);
}
// Stores the sum of order of values
int score = 0, n = 0;
for (int i = 0; i < alpha.Length; i++) {
// Find the score
if (alpha[i] >= 'A' && alpha[i] <= 'Z')
score += alpha[i] - 'A' + 1;
else
score += alpha[i] - 'a' + 1;
}
// Find the single digit sum
while (score > 0 || n > 9) {
if (score == 0) {
score = n;
n = 0;
}
n += score % 10;
score /= 10;
}
// Return the resultant sum
return n;
}
// Driver Code
public static void Main()
{
string S = "GeeksforGeeks";
Console.WriteLine(findTheSum(S));
}
}
// This code is contributed by ukasp.
java 描述语言
<script>
// Javascript program for the above approach
function findTheSum(str) {
let alpha = [];
// Traverse the given string
for (let i = 0; i < str.length; i++) {
// If character is an alphabet
if (
(str[i].charCodeAt(0) >= "A".charCodeAt(0) &&
str[i].charCodeAt(0) <= "Z".charCodeAt(0)) ||
(str[i].charCodeAt(0) >= "a".charCodeAt(0) &&
str[i].charCodeAt(0) <= "z".charCodeAt(0))
)
alpha.push(str[i]);
}
// Stores the sum of order of values
let score = 0,
n = 0;
for (let i = 0; i < alpha.length; i++) {
// Find the score
if (
alpha[i].charCodeAt(0) >= "A".charCodeAt(0) &&
alpha[i].charCodeAt(0) <= "Z".charCodeAt(0)
)
score += alpha[i].charCodeAt(0) - "A".charCodeAt(0) + 1;
else score += alpha[i].charCodeAt(0) - "a".charCodeAt(0) + 1;
}
// Find the single digit sum
while (score > 0 || n > 9) {
if (score == 0) {
score = n;
n = 0;
}
n += score % 10;
score = Math.floor(score / 10);
}
// Return the resultant sum
return n;
}
// Driver Code
let S = "GeeksforGeeks";
document.write(findTheSum(S));
// This code is contributed by saurabh_jaiswal.
</script>
Output:
7
时间复杂度:O(N) T5辅助空间:** O(1)
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