企业策划书ppt_腾云建站官网_沈阳seo_网站优化公司推荐

方法一：

public class Solution {public String reverseWords(String s) {if(s.length()==1&&!s.equals(" ")){return s;}List<String> res=new ArrayList<>();int start=0;for(int i=1;i<s.length();i++){if(s.charAt(i)!=' '&& s.charAt(i - 1)==' '){start=i;}if(s.charAt(i-1)!=' '&&s.charAt(i)==' '){res.add(s.substring(start,i));}if(i==s.length()-1&&s.charAt(i)!=' '){res.add(s.substring(start,i+1));}}Collections.reverse(res);return String.join(" ", res);}
}

class Solution(object):def reverseWords(self, s):if len(s)==1 and s!=" ":return sres=list()start=0for i in range(1,len(s)):if s[i]!=" " and s[i-1]==" ":start=iif s[i-1]!=" " and s[i]==" ":res.append(s[start:i])if i==len(s)-1 and s[i]!=" ":res.append(s[start:i+1])res.reverse()return " ".join(res)

方法二：

public class Solution {public String reverseWords(String s) {// 使用 split 方法将字符串按空格分割成单词数组// 注意：split 方法会处理多余的空格String[] words = s.trim().split("\\s+");// 使用双指针反转单词数组int left = 0, right = words.length - 1;while (left < right) {String temp = words[left];words[left] = words[right];words[right] = temp;left++;right--;}// 使用 String.join 将单词数组拼接成字符串return String.join(" ", words);}
}

class Solution(object):def reverseWords(self, s):# 先反转整个句子，然后反转其中的每个单词# 空间复杂度为O(1)的解法words = s.split()left, right = 0, len(words) - 1while left < right:words[left], words[right] = words[right], words[left]left += 1right -= 1return " ".join(words)