检查给定句子的一组简单语法规则
原文:https://www . geesforgeks . org/check-given-句子-given-set-simple-grammer-rules/
一个简单的句子，如果语法正确，如果它满足给定的规则。以下是给定的规则。 1。句子必须以大写字符开头(如名词/我/我们/他等)。) 2。然后是小写字符。 3。单词之间必须有空格。 4。那么这个句子必须以句号结尾。)后一句话。 5。不允许有两个连续的空格。 6。不允许两个连续的大写字符。 7。但是，句子可以在大写字符后结束。 示例:
Correct sentences -
"My name is Ram."
"The vertex is S."
"I am single."
"I love Geeksquiz and Geeksforgeeks."
Incorrect sentence -
"My name is KG."
"I lovE cinema."
"GeeksQuiz. is a quiz site."
" You are my friend."
"I love cinema"
问题:给定一个句子，根据上述给定规则验证给定句子。 强烈建议尽量减少浏览器，先自己试试这个。 想法是对给定的规则集使用自动机。 算法: 1。检查角箱 …..1.a)检查句子中的第一个字符是否大写。 …..1.b)检查最后一个字符是否是句号。 2。对于字符串的其余部分，这个问题可以通过遵循状态图来解决。请参考下面的状态图。
3.我们需要保持字符串中不同字符的先前和当前状态。基于此，我们总是可以验证每个被遍历字符的句子。 下面是一个基于 C 的实现。(顺便说一下，根据规则和代码，这句话也是正确的)
C++
// C program to validate a given sentence for a set of rules
#include<stdio.h>
#include<string.h>
#include<stdbool.h>
// Method to check a given sentence for given rules
bool checkSentence(char str[])
{
// Calculate the length of the string.
int len = strlen(str);
// Check that the first character lies in [A-Z].
// Otherwise return false.
if (str[0] < 'A' || str[0] > 'Z')
return false;
//If the last character is not a full stop(.) no
//need to check further.
if (str[len - 1] != '.')
return false;
// Maintain 2 states. Previous and current state based
// on which vertex state you are. Initialise both with
// 0 = start state.
int prev_state = 0, curr_state = 0;
//Keep the index to the next character in the string.
int index = 1;
//Loop to go over the string.
while (str[index])
{
// Set states according to the input characters in the
// string and the rule defined in the description.
// If current character is [A-Z]. Set current state as 0.
if (str[index] >= 'A' && str[index] <= 'Z')
curr_state = 0;
// If current character is a space. Set current state as 1.
else if (str[index] == ' ')
curr_state = 1;
// If current character is [a-z]. Set current state as 2.
else if (str[index] >= 'a' && str[index] <= 'z')
curr_state = 2;
// If current state is a dot(.). Set current state as 3.
else if (str[index] == '.')
curr_state = 3;
// Validates all current state with previous state for the
// rules in the description of the problem.
if (prev_state == curr_state && curr_state != 2)
return false;
if (prev_state == 2 && curr_state == 0)
return false;
// If we have reached last state and previous state is not 1,
// then check next character. If next character is '\0', then
// return true, else false
if (curr_state == 3 && prev_state != 1)
return (str[index + 1] == '\0');
index++;
// Set previous state as current state before going over
// to the next character.
prev_state = curr_state;
}
return false;
}
// Driver program
int main()
{
char *str[] = { "I love cinema.", "The vertex is S.",
"I am single.", "My name is KG.",
"I lovE cinema.", "GeeksQuiz. is a quiz site.",
"I love Geeksquiz and Geeksforgeeks.",
" You are my friend.", "I love cinema" };
int str_size = sizeof(str) / sizeof(str[0]);
int i = 0;
for (i = 0; i < str_size; i++)
checkSentence(str[i])? printf("\"%s\" is correct \n", str[i]):
printf("\"%s\" is incorrect \n", str[i]);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to validate a given sentence
// for a set of rules
class GFG
{
// Method to check a given sentence for given rules
static boolean checkSentence(char[] str)
{
// Calculate the length of the string.
int len = str.length;
// Check that the first character lies in [A-Z].
// Otherwise return false.
if (str[0] < 'A' || str[0] > 'Z')
return false;
// If the last character is not a full stop(.)
// no need to check further.
if (str[len - 1] != '.')
return false;
// Maintain 2 states. Previous and current state
// based on which vertex state you are.
// Initialise both with 0 = start state.
int prev_state = 0, curr_state = 0;
// Keep the index to the next character in the string.
int index = 1;
// Loop to go over the string.
while (index <= str.length)
{
// Set states according to the input characters
// in the string and the rule defined in the description.
// If current character is [A-Z]. Set current state as 0.
if (str[index] >= 'A' && str[index] <= 'Z')
curr_state = 0;
// If current character is a space.
// Set current state as 1.
else if (str[index] == ' ')
curr_state = 1;
// If current character is [a-z].
// Set current state as 2.
else if (str[index] >= 'a' && str[index] <= 'z')
curr_state = 2;
// If current state is a dot(.).
// Set current state as 3.
else if (str[index] == '.')
curr_state = 3;
// Validates all current state with previous state
// for the rules in the description of the problem.
if (prev_state == curr_state && curr_state != 2)
return false;
if (prev_state == 2 && curr_state == 0)
return false;
// If we have reached last state and previous state
// is not 1, then check next character. If next character
// is '\0', then return true, else false
if (curr_state == 3 && prev_state != 1)
return (index + 1 == str.length);
index++;
// Set previous state as current state
// before going over to the next character.
prev_state = curr_state;
}
return false;
}
// Driver Code
public static void main(String[] args)
{
String[] str = { "I love cinema.", "The vertex is S.",
"I am single.", "My name is KG.",
"I lovE cinema.", "GeeksQuiz. is a quiz site.",
"I love Geeksquiz and Geeksforgeeks.",
" You are my friend.", "I love cinema" };
int str_size = str.length;
int i = 0;
for (i = 0; i < str_size; i++)
{
if (checkSentence(str[i].toCharArray()))
System.out.println("\"" + str[i] +
"\"" + " is correct");
else
System.out.println("\"" + str[i] +
"\"" + " is incorrect");
}
}
}
// This code is contributed by
// sanjeev2552
计算机编程语言
# Python program to validate a given sentence for a set of rules
# Method to check a given sentence for given rules
def checkSentence(string):
# Calculate the length of the string.
length = len(string)
# Check that the first character lies in [A-Z].
# Otherwise return false.
if string[0] < 'A' or string[0] > 'Z':
return False
# If the last character is not a full stop(.) no
# need to check further.
if string[length-1] != '.':
return False
# Maintain 2 states. Previous and current state based
# on which vertex state you are. Initialise both with
# 0 = start state.
prev_state = 0
curr_state = 0
# Keep the index to the next character in the string.
index = 1
# Loop to go over the string.
while (string[index]):
# Set states according to the input characters in the
# string and the rule defined in the description.
# If current character is [A-Z]. Set current state as 0.
if string[index] >= 'A' and string[index] <= 'Z':
curr_state = 0
# If current character is a space. Set current state as 1.
elif string[index] == ' ':
curr_state = 1
# If current character is a space. Set current state as 2.
elif string[index] >= 'a' and string[index] <= 'z':
curr_state = 2
# If current character is a space. Set current state as 3.
elif string[index] == '.':
curr_state = 3
# Validates all current state with previous state for the
# rules in the description of the problem.
if prev_state == curr_state and curr_state != 2:
return False
# If we have reached last state and previous state is not 1,
# then check next character. If next character is '\0', then
# return true, else false
if prev_state == 2 and curr_state == 0:
return False
# Set previous state as current state before going over
# to the next character.
if curr_state == 3 and prev_state != 1:
return True
index += 1
prev_state = curr_state
return False
# Driver program
string = ["I love cinema.", "The vertex is S.",
"I am single.", "My name is KG.",
"I lovE cinema.", "GeeksQuiz. is a quiz site.",
"I love Geeksquiz and Geeksforgeeks.",
" You are my friend.", "I love cinema"]
string_size = len(string)
for i in xrange(string_size):
if checkSentence(string[i]):
print "\"" + string[i] + "\" is correct"
else:
print "\"" + string[i] + "\" is incorrect"
# This code is contributed by BHAVYA JAIN
C
// C# program to validate a given sentence
// for a set of rules
using System;
class GFG
{
// Method to check a given sentence for given rules
static bool checkSentence(char[] str)
{
// Calculate the length of the string.
int len = str.Length;
// Check that the first character lies in [A-Z].
// Otherwise return false.
if (str[0] < 'A' || str[0] > 'Z')
return false;
// If the last character is not a full stop(.)
// no need to check further.
if (str[len - 1] != '.')
return false;
// Maintain 2 states. Previous and current state
// based on which vertex state you are.
// Initialise both with 0 = start state.
int prev_state = 0, curr_state = 0;
// Keep the index to the next character in the string.
int index = 1;
// Loop to go over the string.
while (index <= str.Length)
{
// Set states according to the input characters
// in the string and the rule defined in the description.
// If current character is [A-Z]. Set current state as 0.
if (str[index] >= 'A' && str[index] <= 'Z')
curr_state = 0;
// If current character is a space.
// Set current state as 1.
else if (str[index] == ' ')
curr_state = 1;
// If current character is [a-z].
// Set current state as 2.
else if (str[index] >= 'a' && str[index] <= 'z')
curr_state = 2;
// If current state is a dot(.).
// Set current state as 3.
else if (str[index] == '.')
curr_state = 3;
// Validates all current state with previous state
// for the rules in the description of the problem.
if (prev_state == curr_state && curr_state != 2)
return false;
if (prev_state == 2 && curr_state == 0)
return false;
// If we have reached last state and previous state
// is not 1, then check next character. If next character
// is '\0', then return true, else false
if (curr_state == 3 && prev_state != 1)
return (index + 1 == str.Length);
index++;
// Set previous state as current state
// before going over to the next character.
prev_state = curr_state;
}
return false;
}
// Driver Code
public static void Main(String[] args)
{
String[] str = { "I love cinema.", "The vertex is S.",
"I am single.", "My name is KG.",
"I lovE cinema.", "GeeksQuiz. is a quiz site.",
"I love Geeksquiz and Geeksforgeeks.",
" You are my friend.", "I love cinema" };
int str_size = str.Length;
int i = 0;
for (i = 0; i < str_size; i++)
{
if (checkSentence(str[i].ToCharArray()))
Console.WriteLine("\"" + str[i] +
"\"" + " is correct");
else
Console.WriteLine("\"" + str[i] +
"\"" + " is incorrect");
}
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript program to validate a given sentence
// for a set of rules
// Method to check a given sentence for given rules
function checkSentence(str) {
// Calculate the length of the string.
var len = str.length;
// Check that the first character lies in [A-Z].
// Otherwise return false.
if (
str[0].charCodeAt(0) < "A".charCodeAt(0) ||
str[0].charCodeAt(0) > "Z".charCodeAt(0)
)
return false;
// If the last character is not a full stop(.)
// no need to check further.
if (str[len - 1] !== ".") return false;
// Maintain 2 states. Previous and current state
// based on which vertex state you are.
// Initialise both with 0 = start state.
var prev_state = 0,
curr_state = 0;
// Keep the index to the next character in the string.
var index = 1;
// Loop to go over the string.
while (index <= str.length) {
// Set states according to the input characters
// in the string and the rule defined in the description.
// If current character is [A-Z]. Set current state as 0.
if (
str[index].charCodeAt(0) >= "A".charCodeAt(0) &&
str[index].charCodeAt(0) <= "Z".charCodeAt(0)
)
curr_state = 0;
// If current character is a space.
// Set current state as 1.
else if (str[index] === " ") curr_state = 1;
// If current character is [a-z].
// Set current state as 2.
else if (
str[index].charCodeAt(0) >= "a".charCodeAt(0) &&
str[index].charCodeAt(0) <= "z".charCodeAt(0)
)
curr_state = 2;
// If current state is a dot(.).
// Set current state as 3.
else if (str[index] === ".") curr_state = 3;
// Validates all current state with previous state
// for the rules in the description of the problem.
if (prev_state === curr_state && curr_state !== 2) return false;
if (prev_state === 2 && curr_state === 0) return false;
// If we have reached last state and previous state
// is not 1, then check next character. If next character
// is '\0', then return true, else false
if (curr_state === 3 && prev_state !== 1)
return index + 1 == str.length;
index++;
// Set previous state as current state
// before going over to the next character.
prev_state = curr_state;
}
return false;
}
// Driver Code
var str = [
"I love cinema.",
"The vertex is S.",
"I am single.",
"My name is KG.",
"I lovE cinema.",
"GeeksQuiz. is a quiz site.",
"I love Geeksquiz and Geeksforgeeks.",
" You are my friend.",
"I love cinema",
];
var str_size = str.length;
var i = 0;
for (i = 0; i < str_size; i++) {
var temp = str[i].split("");
if (checkSentence(temp))
document.write('"' + str[i] + '"' + " is correct" + "<br>");
else document.write('"' + str[i] + '"' + " is incorrect" + "<br>");
}
</script>
输出:
"I love cinema." is correct
"The vertex is S." is correct
"I am single." is correct
"My name is KG." is incorrect
"I lovE cinema." is incorrect
"GeeksQuiz. is a quiz site." is incorrect
"I love Geeksquiz and Geeksforgeeks." is correct
" You are my friend." is incorrect
"I love cinema" is incorrect
时间复杂度–O(n)，最坏的情况，因为我们必须遍历整个句子，其中 n 是句子的长度。 辅助空间–O(1) 本文由库马尔·高塔姆供稿。如果你发现任何不正确的地方，或者你想分享更多关于上面讨论的话题的信息，请写评论。
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