所有给定句子中出现的单词数
原文：https://www.geeksforgeeks.org/count-of-words-that-are-present-in-all-the-given-sentences/
给定n个句子。 任务是计算所有这些句子中出现的单词数。 请注意，每个单词仅由小写英文字母组成。
示例：
输入：
arr[] = { "there is a cow", "cow is our mother", "cow gives us milk and milk is sweet", "there is a boy who loves cow" }输出：2
在所有句子中只有单词
"is"和"cow"出现。输入：
arr [] = { "abc aac abcd ccc", "ac aa abc cca", "abca aaac abcd ccc" }输出：0
朴素的方法：朴素的方法是将第一句的每个单词都包含在所有行中，然后将其与其余行进行比较，然后增加计数。
有效方法：对于第一个句子的所有单词，我们可以使用映射来恒定时间检查它是否也存在于所有其他句子中。 将第一个句子的所有单词存储在映射中，并检查所有其他句子中有多少个存储的单词。
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of common words
// in all the sentences
int commonWords(vector<string> S)
{
int m, n, i, j;
// To store separate words
string str;
// It will be used to check if a word is present
// in a particuler string
unordered_map<string, bool> has;
// To store all the words of first string
vector<pair<string, bool> > ans;
pair<string, bool> tmp;
// m will store number of strings in given vector
m = S.size();
i = 0;
// Extract all words of first string and store it in ans
while (i < S[0].size()) {
str = "";
while (i < S[0].size() && S[0][i] != ' ') {
str += S[0][i];
i++;
}
// Increase i to get at starting index
// of the next word
i++;
// If str is not empty store it in map
if (str != "") {
tmp = make_pair(str, true);
ans.push_back(tmp);
}
}
// Start from 2nd line check if any word of
// the first string did not match with
// some word in the current line
for (j = 1; j < m; j++) {
has.clear();
i = 0;
while (i < S[j].size()) {
str = "";
while (i < S[j].size() && S[j][i] != ' ') {
str += S[j][i];
i++;
}
i++;
if (str != "") {
has[str] = true;
}
}
// Check all words of this vector
// if it is not present in current line
// make it false
for (int k = 0; k < ans.size(); k++) {
if (ans[k].second != false
&& has[ans[k].first] == false) {
ans[k].second = false;
}
// This line is used to consider only distinct words
else if (ans[k].second != false
&& has[ans[k].first] == true) {
has[ans[k].first] = false;
}
}
}
// This function will print
// the count of common words
int cnt = 0;
// If current word appears in all the sentences
for (int k = 0; k < ans.size(); k++) {
if (ans[k].second == true)
cnt++;
}
return cnt;
}
// Driver code
int main()
{
vector<string> S;
S.push_back("there is a cow");
S.push_back("cow is our mother");
S.push_back("cow gives us milk and milk is sweet");
S.push_back("there is a boy who loves cow");
cout << commonWords(S);
return 0;
}
Java
// Java implementation of the approach
import java.util.HashMap;
class GFG
{
// Function to return the count of
// common words in all the sentences
static int commonWords(String[] s)
{
int m, i, j;
// To store separate words
String str;
// It will be used to check if a word
// is present in a particuler string
HashMap<String, Boolean> has = new HashMap<>();
// To store all the words of first string
String[] ans1 = new String[100];
boolean[] ans2 = new boolean[100];
int track = 0;
// m will store number of strings
// in given vector
m = s.length;
i = 0;
// Extract all words of first string
// and store it in ans
while (i < s[0].length())
{
str = "";
while (i < s[0].length() &&
s[0].charAt(i) != ' ')
{
str += s[0].charAt(i);
i++;
}
// Increase i to get at starting index
// of the next word
i++;
// If str is not empty store it in map
if (str.compareTo("") != 0)
{
ans1[track] = str;
ans2[track] = true;
track++;
}
}
// Start from 2nd line check if any word of
// the first string did not match with
// some word in the current line
for (j = 1; j < m; j++)
{
has.clear();
i = 0;
while (i < s[j].length())
{
str = "";
while (i < s[j].length() &&
s[j].charAt(i) != ' ')
{
str += s[j].charAt(i);
i++;
}
i++;
if (str.compareTo("") != 0)
has.put(str, true);
}
// Check all words of this vector
// if it is not present in current line
// make it false
for (int k = 0; k < track; k++)
{
// System.out.println(has.get(ans1[k]));
if (ans2[k] != false &&
!has.containsKey(ans1[k]))
ans2[k] = false;
// This line is used to consider
// only distinct words
else if (ans2[k] != false &&
has.containsKey(ans1[k]) &&
has.get(ans1[k]) == true)
has.put(ans1[k], false);
}
}
// This function will print
// the count of common words
int cnt = 0;
// If current word appears
// in all the sentences
for (int k = 0; k < track; k++)
if (ans2[k] == true)
cnt++;
return cnt;
}
// Driver Code
public static void main(String[] args)
{
String[] s = { "there is a cow", "cow is our mother",
"cow gives us milk and milk is sweet",
"there is a boy who loves cow" };
System.out.println(commonWords(s));
}
}
// This code is contributed by
// sanjeev2552
Python3
# Python3 implementation of the approach
from collections import defaultdict
# Function to return the count of
# common words in all the sentences
def commonWords(S):
# It will be used to check if a word
# is present in a particuler string
has = defaultdict(lambda:False)
# To store all the words of first string
ans = []
# m will store number of strings
# in given vector
m = len(S)
i = 0
# Extract all words of first string
# and store it in ans
while i < len(S[0]):
string = ""
while i < len(S[0]) and S[0][i] != ' ':
string += S[0][i]
i += 1
# Increase i to get at starting
# index of the next word
i += 1
# If str is not empty store it in map
if string != "":
ans.append([string, True])
# Start from 2nd line check if any word
# of the first string did not match with
# some word in the current line
for j in range(1, m):
has.clear()
i = 0
while i < len(S[j]):
string = ""
while i < len(S[j]) and S[j][i] != ' ':
string += S[j][i]
i += 1
i += 1
if string != "":
has[string] = True
# Check all words of this vector
# if it is not present in current
# line make it false
for k in range(0, len(ans)):
if (ans[k][1] != False and
has[ans[k][0]] == False):
ans[k][1] = False
# This line is used to consider
# only distinct words
elif (ans[k][1] != False
and has[ans[k][0]] == True):
has[ans[k][0]] = False
# This function will print
# the count of common words
cnt = 0
# If current word appears in all
# the sentences
for k in range(0, len(ans)):
if ans[k][1] == True:
cnt += 1
return cnt
# Driver code
if __name__ == "__main__":
S = []
S.append("there is a cow")
S.append("cow is our mother")
S.append("cow gives us milk and milk is sweet")
S.append("there is a boy who loves cow")
print(commonWords(S))
# This code is contributed by Rituraj Jain
输出：
2
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