给定一个字符串,请你把字符串中所有用空格隔开的单词翻转,保留原有的空格和单词顺序
数据范围:字符串长度满足 , 字符串中仅包含小写英文字母和空格
package main import "strings" /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param str string字符串 * @return string字符串 */ func reverseWord( str string ) string { arr:=strings.Split(str," ") for i,s:=range arr{ arr[i]=reverse(s) } return strings.Join(arr," ") } func reverse(s string)string{ arr:=strings.Split(s,"") i,j:=0,len(s)-1 for i<j{ arr[i],arr[j]=arr[j],arr[i] i++ j-- } return strings.Join(arr,"") }
class Solution: def reverseWord(self, str: str) -> str: # write code here # Solution1:只能保证单词间只存在一个空格的情况 # m = re.findall(r'\w+',str) # count = len(m) # for i in range(0,count): # m[i] = m[i][::-1] # return " ".join(m) res = [] # 存储输出后的字符串 tempStr = "" # 零时字符串对单词进行处理 n = len(str) # 字符串长度 for i in range(0, n): if str[i] != " ": # 当当前字符为字母时,存入零食变量tempStr tempStr += str[i] if i == n - 1: tempStr = tempStr[::-1] res.append(tempStr) else: tempStr = tempStr[::-1] res.append(tempStr) tempStr = "" res.append(str[i]) return "".join(res)