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);
    }
}