LeetCode（String)1816. Truncate Sentence

叁歲伎倆 2023-09-27 20:52 91阅读 0赞

# 1.问题 #

A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each of the words consists of only uppercase and lowercase English letters (no punctuation).

For example, “Hello World”, “HELLO”, and “hello world hello world” are all sentences.  
You are given a sentence s and an integer k. You want to truncate s such that it contains only the first k words. Return s after truncating it.

Example 1:

> Input: s = “Hello how are you Contestant”, k = 4  
> Output: “Hello how are you”  
> Explanation:  
> The words in s are \[“Hello”, “how” “are”, “you”, “Contestant”\].  
> The first 4 words are \[“Hello”, “how”, “are”, “you”\].  
> Hence, you should return “Hello how are you”.

Example 2:

> Input: s = “What is the solution to this problem”, k = 4  
> Output: “What is the solution”  
> Explanation:  
> The words in s are \[“What”, “is” “the”, “solution”, “to”, “this”, “problem”\].  
> The first 4 words are \[“What”, “is”, “the”, “solution”\].  
> Hence, you should return “What is the solution”.

Example 3:

> Input: s = “chopper is not a tanuki”, k = 5  
> Output: “chopper is not a tanuki”

Constraints:

1 <= s.length <= 500  
k is in the range \[1, the number of words in s\].  
s consist of only lowercase and uppercase English letters and spaces.  
The words in s are separated by a single space.  
There are no leading or trailing spaces.

# 2. 解题思路 #

## 方法1： ##

> 1.Java 中的 `split()`方法将字符串 s 按照空格分割成字符串数组  
> 2. 然后取前 k 个字符串拼接成一个新的字符串即可

## 方法2： ##

> 双指针  
> 1.用指针 i 和 j 分别指向字符串 s 的开始位置和结束位置  
> 2. 然后从前往后扫描字符串 s，遇到空格就将 j 指针更新为当前位置，直到找到前 k 个单词为止  
> 3. 最后截取字符串 s 的开始位置到 j 指针位置的子串作为输出结果。

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public String truncateSentence(String s, int k) {
        
            String[] splited = s.split(" "); // 将 s 按照空格分割成字符串数组
            String result = "";
            for(int i=0;i<=k-1;i++){
         // 取前 k 个字符串拼接成新的字符串
                result+=splited[i]+" "; // 在字符串之间添加一个空格
            }
            return result.trim(); // 去掉字符串首尾的空格并返回
        }
    }

> 解题思路基本相同，区别在使用字符串拼接。
> 
>  *  比使用 StringBuilder 或 StringBuffer 的 append() 方法要低，因为每次字符串拼接都会创建一个新的字符串对象。

class Solution {
        
        public String truncateSentence(String s, int k) {
        
            
            String[] arr = s.split(" ");// 将 s 按照空格分割成字符串数组
            StringBuilder sb = new StringBuilder();// 在字符串之间添加一个空格
            
            for(int i=0;i<k;i++) {
        // 取前 k 个字符串拼接成新的字符串
                sb.append(arr[i]);
                sb.append(" ");
            }
            return sb.toString().trim();// 去掉字符串首尾的空格并返回
        }
    }

> 解题思路基本相同，`区别在最后去掉字符串首尾的空格并返回`

class Solution {
        
        public String truncateSentence(String s, int k) {
        
            String[] words = s.split(" "); // 将 s 按照空格分割成字符串数组
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < k; i++) {
         // 取前 k 个字符串拼接成新的字符串
                sb.append(words[i]);
                if (i != k - 1) {
         // 如果不是最后一个字符串，则在后面添加一个空格
                    sb.append(" ");
                }
            }
            return sb.toString();
        }
    }

## 代码2： ##

class Solution {
        
        public String truncateSentence(String s, int k) {
        
            int i = 0, j = 0, count = 0;
            while (j < s.length() && count < k) {
        
                if (s.charAt(j) == ' ') {
        
                    count++;
                }
                j++;
            }
            if (count == k && j < s.length()) {
         // 如果找到前 k 个单词，并且 j 指针不在字符串末尾
                return s.substring(i, j - 1); // 返回子串，不包括 j 指针所在位置的空格
            } else {
         // 否则，返回整个字符串 s
                return s;
            }
        }
    }