给定一个字符串,你需要反转字符串中每个单词的字符顺序,同时仍保留空格和单词的初始顺序。
示例:
输入:"Let's take LeetCode contest" 输出:"s'teL ekat edoCteeL tsetnoc"
提示:
- 在字符串中,每个单词由单个空格分隔,并且字符串中不会有任何额外的空格。
class Solution:
def reverseWords(self, s: str) -> str:
return ' '.join([t[::-1] for t in s.split(' ')])
class Solution {
public String reverseWords(String s) {
StringBuilder res = new StringBuilder();
for (String t : s.split(" ")) {
for (int i = t.length() - 1; i >= 0; --i) {
res.append(t.charAt(i));
}
res.append(" ");
}
return res.substring(0, res.length() - 1);
}
}
class Solution {
public:
string reverseWords(string s) {
for (int i = 0, n = s.size(); i < n; ++i)
{
int j = i;
while (++j < n && s[j] != ' ');
reverse(s.begin() + i, s.begin() + j);
i = j;
}
return s;
}
};
func reverseWords(s string) string {
t := []byte(s)
for i := 0; i < len(t); i++ {
j := i
for j < len(t) && t[j] != ' ' {
j++
}
for st, ed := i, j-1; st < ed; st, ed = st+1, ed-1 {
t[st], t[ed] = t[ed], t[st]
}
i = j
}
return string(t)
}