所有给定句子中出现的单词数

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。
例子：

Input: arr[] = {
“there is a cow”,
“cow is our mother”,
“cow gives us milk and milk is sweet”,
“there is a boy who loves cow”}
Output: 2
Only the words “is” and “cow” appear in all of the sentences.
Input: arr[] = {
“abc aac abcd ccc”,
“ac aa abc cca”,
“abca aaac abcd ccc”}
Output: 0

编程需要懂一点英语

朴素的方法：朴素的方法是将第一句话的每个单词与其余行进行比较，如果它也出现在所有行中，则增加计数。
高效的方法：对于第一句话的所有单词，我们可以通过使用地图检查它是否也在恒定时间内出现在所有其他句子中。将第一个句子的所有单词存储在映射中，并检查这些存储的单词中有多少存在于所有其他句子中。

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count of common words
// in all the sentences
int commonWords(vector 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 particular string
    unordered_map has;
 
    // To store all the words of first string
    vector > ans;
 
    pair 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 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 particular string
        HashMap 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 particular 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

C#

// C# implementation of the
// above approach
using System;
using System.Collections.Generic;
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
  // particular string
  Dictionary has =
             new Dictionary();
 
  // To store all the words of
  // first string
  String[] ans1 = new String[100];
  bool[] ans2 = new bool[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][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.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][i] != ' ')
      {
        str += s[j][i];
        i++;
      }
 
      i++;
      if (str.CompareTo("") != 0)
        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 < track; k++)
    {
      // Console.WriteLine(has[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[ans1[k]] == true)
        has[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" };
  Console.WriteLine(commonWords(s));
}
}
 
// This code is contributed by Rajput-Ji

输出：

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