Given an input string, reverse the string word by word.
Example 1:
Input: "the sky is blue"
Output: "blue is sky the"
Example 2:
Input: " hello world! "
Output: "world! hello"
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:
Input: "a good example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
Note:
A word is defined as a sequence of non-space characters.
Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
You need to reduce multiple spaces between two words to a single space in the reversed string.
Follow up:
For C programmers, try to solve it in-place in O(1) extra space.
string reverseWords(string s) { // time: O(n); space: O(n)
istringstream iss(s);
string buff, res;
iss >> res;
while (iss >> buff) {
res = buff + ' ' + res;
}
if (!res.empty() && res[0] == ' ') res.clear(); // handle corner case such as s = " "
return res;
}
string reverseWords(string s) { // time: O(n); space: O(n)
string res;
int n = s.length();
for (int i = 0; i < n; ++i) {
if (s[i] == ' ') continue;
int word_start = i;
while (i < n && s[i] != ' ') {
++i;
}
// add space if res is not empty
if (!res.empty()) {
res = ' ' + res;
}
res = s.substr(word_start, i - word_start) + res;
}
return res;
}
// In-Place Algorithm
string reverseWords(string s) { // time: O(n); space: O(1)
reverse(s.begin(), s.end());
int n = s.length(), storeIdx = 0;
for (int i = 0; i < n; ++i) {
if (s[i] == ' ') continue;
// add space if res is not empty
if (storeIdx > 0) s[storeIdx++] = ' ';
int j = i;
while (j < n && s[j] != ' ') {
s[storeIdx++] = s[j++];
}
reverse(s.begin() + storeIdx - (j - i), s.begin() + storeIdx);
i = j;
}
s.resize(storeIdx);
return s;
}