将一个句子中出现的所有单词精确打印 K 次
原文:https://www . geesforgeks . org/print-all-words-in-a-in-in-a-their-k-times/
给定一个由小写字母和整数组成的字符串 S ，任务是打印字符串 S 中出现 K 次的所有单词。
示例:
输入: S =“香蕉是黄色的，太阳花也是黄色的”，K = 2 输出:“是”“黄色的”“在” 解释:单词“是”“黄色的”“在”在字符串中出现两次。
输入: S =“极客为极客”，K = 2 输出:“极客”
方法:按照以下步骤解决问题:
下面是上述方法的实现:
C++
// CPP program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to print all the words
// occurring k times in a string
void kFreqWords(string S, int K)
{
// Stores the words
string temp = "";
vector<string> l;
for (auto x: S)
{
if(x == ' ')
{
l.push_back(temp);
temp = "";
}
else
temp += x;
}
// Traverse the list
for (auto x: l)
{
// Check for count
if (count(l.begin(), l.end(), x) == K)
{
// Print the word
cout << x << endl;
// Remove from list
remove(l.begin(),l.end(), x);
}
}
}
// Driver Code
int main()
{
// Given string
string S = "banana is in yellow and sun flower is also in yellow ";
// Given value of K
int K = 2;
// Function call to find
// all words occurring K times
kFreqWords(S, K);
}
// This code is contributed by SURENDRA_GANGWAR.
Python 3
# Python3 program for the above approach
# Function to print all the words
# occurring k times in a string
def kFreqWords(S, K):
# Stores the words
l = list(S.split(" "))
# Traverse the list
for i in l:
# Check for count
if l.count(i) == K:
# Print the word
print(i)
# Remove from list
l.remove(i)
# Driver Code
if __name__ == "__main__":
# Given string
S = "banana is in yellow and sun flower is also in yellow"
# Given value of K
K = 2
# Function call to find
# all words occurring K times
kFreqWords(S, K)
Output
is
yellow
in
时间复杂度:O(N) T5辅助空间:** O(1)
方法 2:使用内置的 python 函数:
- 因为一个句子中的所有单词都用空格隔开。
- 我们必须用 split()将句子用空格分开。
- 我们将所有的单词用空格分开，并存储在一个列表中。
- 使用计数器功能统计单词的出现频率
- 遍历频率字典，打印频率为 k 的单词
下面是上述方法的实现:
Python 3
# Python program for the above approach
from collections import Counter
# Python program to print words
# which occurs k times
def printWords(sentence, k):
# splitting the string
lis = list(sentence.split(" "))
# Calculating frequency of every word
frequency = Counter(lis)
# Traversing the frequency
for i in frequency:
# checking if frequency is k
if(frequency[i] == k):
# print the word
print(i, end=" ")
# Driver code
# Given string
sentence = "sky is blue and my favourite color is blue"
# Given value of K
K = 2
printWords(sentence, K)
# this code is contributed by vikkycirus
java 描述语言
<script>
// JavaScript program for the above approach
/*
Function to print words
which occur k times
*/
function printWords(sentence, k)
{
// splitting the string
let lis = sentence.split(" ");
// Calculating frequency of each word
frequency = {};
for(const word of lis)
{
// if word exists, increase the value by 1, otherwise mark it as 1
frequency[word] = frequency[word] ? frequency[word] + 1 : 1;
}
// Iterating through frequency
for(const key of Object.keys(frequency)){
// if frequency is k
if(frequency[key] == k){
// print the word
document.write(key + " ");
}
}
}
// Driver code
// Given string
let sentence = "sky is blue and my favourite color is blue"
// Given value of K
let K = 2
printWords(sentence, K)
// This code is contributed by mostaptname
</script>
输出:
is blue
时间复杂度: O(N)
空间复杂度: O(N)
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