LeetCode（Sorting)1859. Sorting the Sentence

女爷i 2023-09-23 23:52 103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 word consists of lowercase and uppercase English letters.

A sentence can be shuffled by appending the 1-indexed word position to each word then rearranging the words in the sentence.

For example, the sentence “This is a sentence” can be shuffled as “sentence4 a3 is2 This1” or “is2 sentence4 This1 a3”.  
Given a shuffled sentence s containing no more than 9 words, reconstruct and return the original sentence.

Example 1:

> Input: s = “is2 sentence4 This1 a3”  
> Output: “This is a sentence”  
> Explanation: Sort the words in s to their original positions “This1 is2 a3 sentence4”, then remove the numbers.

Example 2:

> Input: s = “Myself2 Me1 I4 and3”  
> Output: “Me Myself and I”  
> Explanation: Sort the words in s to their original positions “Me1 Myself2 and3 I4”, then remove the numbers.

Constraints:

*  2 <= s.length <= 200
 *  s consists of lowercase and uppercase English letters, spaces, and digits from 1 to 9.
 *  The number of words in s is between 1 and 9.
 *  The words in s are separated by a single space.
 *  s contains no leading or trailing spaces.

# 2. 解题思路 #

## 方法1： ##

> 1.s去掉空白，转成数组s1;新建数组str，长度为s1;定义一个空的String为result  
> 2.遍历数组s1,将元素中的最后索引的数值，放入到str数组中，取从0到数值之前的字符  
> 3.数组的元素遍历，并添加到result  
> 4.返回result

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public String sortSentence(String s) {
        
            String[] s1 = s.split(" ");//1.s去掉空白，转成数组s1;新建数组str，长度为s1;
            String[] str=new String[s1.length];
    
            for(int i=0;i<s1.length;i++){
        //2.遍历数组s1,将元素中的最后索引的数值，放入到str数组中，取从0到数值之前的字符
                int index= s1[i].length()-1;
                int nums= Character.getNumericValue( s1[i].charAt(index));
                // int index = Integer.parseInt (String.valueOf ( s1[i].charAt (index));
                //或者不需要转成int，char也是可以
                //char nums=  s1[i].charAt(index);str[nums-'1']= s1[i].substring(0,index)
                str[nums-1]= s1[i].substring(0,index);
            }
    
            return String.join(" ", str);//4.使用join（）插入空白，返回str
            
        }
    }

解题思路基本相同，不同的在于for循环遍历放入到String result中，把最后的的空格用trim（）方法去除

class Solution {
        
        public String sortSentence(String s) {
        
            String[] s1 = s.split(" ");//1.s去掉空白，转成数组s1;新建数组str，长度为s1，定义String的result为空;
            String[] str=new String[s1.length];
            String result = "";
    
            for(int i=0;i<s1.length;i++){
        //2.遍历数组s1,将元素中的最后索引的数值，放入到str数组中，取从0到数值之前的字符
                int index= s1[i].length()-1;
                int nums= Character.getNumericValue( s1[i].charAt(index));
                // int index = Integer.parseInt (String.valueOf ( s1[i].charAt (index));
                //或者不需要转成int，char也是可以
                //char nums=  s1[i].charAt(index);str[nums-'1']= s1[i].substring(0,index)
                str[nums-1]= s1[i].substring(0,index);
            }
    
            for (int i = 0; i < str.length; i++) {
        
               result += str[i] + " ";
            }
    
            return result.trim();//4.返回result
            
        }
    }

## 代码2： ##

> 和代码1解题思路基本相同

class Solution {
        
        public String sortSentence(String s) {
        
            String []ans = new String[s.split(" ").length];
            for(String st: s.split(" ")){
        
                ans[st.charAt(st.length()-1) - '1'] = st.substring(0,st.length()-1);
            }
            return String.join(" ", ans);
            
        }
    }