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 1:
输入: "Let's take LeetCode contest"
输出: "s'teL ekat edoCteeL tsetnoc"
注意: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 {
public:
string reverseWords(string s) {
string result;
int first = 0;
int second = 0;
for(int i = 0; i < s.size(); i++){
if(s[i] == ' '){
second = i;
string temp = s.substr(first,second-first);
temp = singleWord(temp);
result += temp;
result += ' ';
first = i+1;
}
if(i == s.size()-1){
second = s.size();
string temp = s.substr(first, second-first);
temp = singleWord(temp);
result += temp;
}
}
return result;
}
//反转单个单词
string singleWord(string& s){
for(int i =0; i < s.size()/2; i++)
{
char temp = s[s.size()-1-i];
s[s.size()-1-i] = s[i];
s[i] = temp;
}
return s;
}
};
看到有的人使用了字符串的reverse函数
class Solution {
public:
string reverseWords(string s) {
s = s + ' ';
auto begin = s.begin(), end = s.begin();
while(end != s.end()) {
if(*end == ' ') {
reverse(begin, end);
begin = end+1;
}
++end;
}
s.erase(end-1);
return s;
}
};
//作者:pu-shang-qing-feng