检查两个单词的总和是否等于目标单词
原文:https://www . geesforgeks . org/check-if-summary-of-two-word-等于-target-word/
给定三个大小分别为 L、M 和 N 的字符串AB和 C ，并且仅由小于“K”的小写英文字母组成。任务是通过将字母与字母列表中的索引值映射并串联起来，将字符串解码成整数后，检查字符串 A 和 B 的和是否等于字符串 C 。
示例:
输入: A = "acb "，B = "cba "，C = " CDB " T3】输出:是 T6】说明:
- 在用字母列表中的索引值(即 0、1 和 2)替换字符“A”、“b”和“c”后，字符串 A 修改为整数 021。
- 字符串 B 在用字母列表中的索引值(即 0、1 和 2)替换字符“a”、“B”和“c”后，修改为整数 210。
- 在用字母列表中的索引值(即 1、2 和 3)替换字符“b”、“C”和“d”后，字符串 C 修改为整数 231。
字符串 A 和 B 的和(即 21+210 = 231)等于 231，是字符串 c 的值，因此打印“是”。
输入:A =“AAA”，B =“BCB”，C =“BCA” T3】输出:否
方法:这个问题可以使用类似的方法来解决，该方法用于寻找表示为字符串的两个大数的和。按照以下步骤解决问题:
- 反转琴弦T2AB和 C 。
- 初始化两个变量，将 curr 和 rem 设为 0 ，将值存储在 i th 位置以及字符串 A 和 B 之和的剩余部分。
- 使用变量 i 迭代范围【0，最大值(L，最大值(M，N))】，并执行以下步骤:
- 在变量 curr 中存储字符串 A 和 B 的IthT3】索引处的字符总和。
- 将 curr 更新为 curr = curr+rem ，然后将 rem 更新为 rem = curr/10。
- 现在检查 i 是否小于 N 和 curr%10 是否不等于字符串 C 的C【I】-即 i 第个字符处的值，然后打印“ No ”和 return 。
- 最后，完成上述步骤后，如果 rem 大于 0 ，则打印“否”。否则，打印“是”。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check weather summation
// of two words equal to target word
string isSumEqual(string A, string B, string C)
{
// Store the length of each string
int L = A.length();
int M = B.length();
int N = A.length();
// Reverse the strings A, B and C
reverse(A.begin(), A.end());
reverse(B.begin(), B.end());
reverse(C.begin(), C.end());
// Stores the remainder
int rem = 0;
// Iterate in the range
// [0, max(L, max(M, N))]
for (int i = 0; i < max(L, max(M, N)); i++) {
// Stores the integer at ith
// position from the right in
// the sum of A and B
int curr = rem;
// If i is less than L
if (i < L)
curr += A[i] - 'a';
// If i is less than M
if (i < M)
curr += B[i] - 'a';
// Update rem and curr
rem = curr / 10;
curr %= 10;
// If i is less than N
// and curr is not equal
// to C[i]-'a', return "No"
if (i < N && curr != C[i] - 'a') {
return "No";
}
}
// If rem is greater
// than 0, return "No"
if (rem)
return "No";
// Otherwise, return "Yes"
else
return "Yes";
}
// Driver Code
int main()
{
// Given Input
string A = "acb", B = "cba", C = "cdb";
// Function Call
cout << isSumEqual(A, B, C);
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{
// Function to check weather summation
// of two words equal to target word
static String isSumEqual(String A, String B, String C)
{
// Store the length of each String
int L = A.length();
int M = B.length();
int N = A.length();
// Reverse the Strings A, B and C
A = reverse(A);
B = reverse(B);
C = reverse(C);
// Stores the remainder
int rem = 0;
// Iterate in the range
// [0, Math.max(L, Math.max(M, N))]
for (int i = 0; i < Math.max(L, Math.max(M, N)); i++) {
// Stores the integer at ith
// position from the right in
// the sum of A and B
int curr = rem;
// If i is less than L
if (i < L)
curr += A.charAt(i) - 'a';
// If i is less than M
if (i < M)
curr += B.charAt(i) - 'a';
// Update rem and curr
rem = curr / 10;
curr %= 10;
// If i is less than N
// and curr is not equal
// to C[i]-'a', return "No"
if (i < N && curr != C.charAt(i) - 'a') {
return "No";
}
}
// If rem is greater
// than 0, return "No"
if (rem>0)
return "No";
// Otherwise, return "Yes"
else
return "Yes";
}
static String reverse(String input) {
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver Code
public static void main(String[] args)
{
// Given Input
String A = "acb", B = "cba", C = "cdb";
// Function Call
System.out.print(isSumEqual(A, B, C));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program for the above approach
# Function to check weather summation
# of two words equal to target word
def isSumEqual(A, B, C):
# Store the length of each string
L = len(A)
M = len(B)
N = len(A)
# Reverse the strings A, B and C
A = A[::-1]
B = B[::-1]
C = C[::-1]
# Stores the remainder
rem = 0
# Iterate in the range
# [0, max(L, max(M, N))]
for i in range(max(L, max(M, N))):
# Stores the integer at ith
# position from the right in
# the sum of A and B
curr = rem
# If i is less than L
if (i < L):
curr += ord(A[i]) - ord('a')
# If i is less than M
if (i < M):
curr += ord(B[i]) - ord('a')
# Update rem and curr
rem = curr // 10
curr %= 10
# If i is less than N
# and curr is not equal
# to C[i]-'a', return "No"
if (i < N and curr != ord(C[i]) - ord('a')):
return "No"
# If rem is greater
# than 0, return "No"
if (rem):
return "No"
# Otherwise, return "Yes"
else:
return "Yes"
# Driver Code
if __name__ == '__main__':
# Given Input
A = "acb"
B = "cba"
C = "cdb"
# Function Call
print (isSumEqual(A, B, C))
# This code is contributed by mohit kumar 29.
C
// C# program for the above approach
using System;
public class GFG{
// Function to check weather summation
// of two words equal to target word
static String isSumEqual(String A, String B, String C)
{
// Store the length of each String
int L = A.Length;
int M = B.Length;
int N = A.Length;
// Reverse the Strings A, B and C
A = reverse(A);
B = reverse(B);
C = reverse(C);
// Stores the remainder
int rem = 0;
// Iterate in the range
// [0, Math.Max(L, Math.Max(M, N))]
for (int i = 0; i < Math.Max(L, Math.Max(M, N)); i++) {
// Stores the integer at ith
// position from the right in
// the sum of A and B
int curr = rem;
// If i is less than L
if (i < L)
curr += A[i] - 'a';
// If i is less than M
if (i < M)
curr += B[i] - 'a';
// Update rem and curr
rem = curr / 10;
curr %= 10;
// If i is less than N
// and curr is not equal
// to C[i]-'a', return "No"
if (i < N && curr != C[i] - 'a') {
return "No";
}
}
// If rem is greater
// than 0, return "No"
if (rem>0)
return "No";
// Otherwise, return "Yes"
else
return "Yes";
}
static String reverse(String input) {
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("",a);
}
// Driver Code
public static void Main(String[] args)
{
// Given Input
String A = "acb", B = "cba", C = "cdb";
// Function Call
Console.Write(isSumEqual(A, B, C));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript program for the above approach
// Function to check weather summation
// of two words equal to target word
function isSumEqual(A, B, C) {
// Store the length of each string
let L = A.length;
let M = B.length;
let N = A.length;
// Reverse the strings A, B and C
A.split("").reverse().join("");
B.split("").reverse().join("");
C.split("").reverse().join("");
// Stores the remainder
let rem = 0;
// Iterate in the range
// [0, max(L, max(M, N))]
for (let i = 0; i < Math.max(L, Math.max(M, N)); i++) {
// Stores the integer at ith
// position from the right in
// the sum of A and B
let curr = rem;
// If i is less than L
if (i < L)
curr += A[i].charCodeAt(0) - 'a'.charCodeAt(0);
// If i is less than M
if (i < M)
curr += B[i].charCodeAt(0) - 'a'.charCodeAt(0);
// Update rem and curr
rem = Math.floor(curr / 10);
curr %= 10;
// If i is less than N
// and curr is not equal
// to C[i]-'a', return "No"
if (i < N && curr != C[i].charCodeAt(0) -
'a'.charCodeAt(0)) {
return "No";
}
}
// If rem is greater
// than 0, return "No"
if (rem)
return "No";
// Otherwise, return "Yes"
else
return "Yes";
}
// Driver Code
// Given Input
let A = "acb", B = "cba", C = "cdb";
// Function Call
document.write(isSumEqual(A, B, C));
</script>
Output
Yes
时间复杂度:O(L+M+N) T5辅助空间:** O(1)
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