Description#
Given an input string s, reverse the order of the words.
A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.
Return a string of the words in reverse order concatenated by a single space.
Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.
Example 1:
Input: s = "the sky is blue"
Output: "blue is sky the"
Example 2:
Input: s = " hello world "
Output: "world hello"
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:
Input: s = "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.
Constraints:
1 <= s.length <= 104s contains English letters (upper-case and lower-case), digits, and spaces ' '.- There is at least one word in
s.
Follow-up: If the string data type is mutable in your language, can you solve it in-place with O(1) extra space?
Solutions#
Solution 1: Use Language Built-in Functions#
We split the string into a list of strings by spaces, then reverse the list, and finally join the list into a string separated by spaces.
Time complexity $O(n)$, space complexity $O(n)$, where $n$ is the length of the string.
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| class Solution:
def reverseWords(self, s: str) -> str:
return ' '.join(reversed(s.split()))
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| class Solution {
public String reverseWords(String s) {
List<String> words = Arrays.asList(s.trim().split("\\s+"));
Collections.reverse(words);
return String.join(" ", words);
}
}
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| class Solution {
public:
string reverseWords(string s) {
int i = 0;
int j = 0;
int n = s.size();
while (i < n) {
while (i < n && s[i] == ' ') {
++i;
}
if (i < n) {
if (j != 0) {
s[j++] = ' ';
}
int k = i;
while (k < n && s[k] != ' ') {
s[j++] = s[k++];
}
reverse(s.begin() + j - (k - i), s.begin() + j);
i = k;
}
}
s.erase(s.begin() + j, s.end());
reverse(s.begin(), s.end());
return s;
}
};
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| func reverseWords(s string) string {
words := strings.Split(s, " ")
var ans []string
for i := len(words) - 1; i >= 0; i-- {
if words[i] != "" {
ans = append(ans, words[i])
}
}
return strings.Join(ans, " ")
}
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| function reverseWords(s: string): string {
return s.trim().split(/\s+/).reverse().join(' ');
}
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| impl Solution {
pub fn reverse_words(s: String) -> String {
s.split_whitespace().rev().collect::<Vec<&str>>().join(" ")
}
}
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| public class Solution {
public string ReverseWords(string s) {
return string.Join(" ", s.Trim().Split(" ").Where(word => !string.IsNullOrEmpty(word) && !string.IsNullOrEmpty(word.Trim())).Reverse());
}
}
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Solution 2: Two Pointers#
We can use two pointers $i$ and $j$, each time we find a word, add it to the result list, then reverse the result list, and finally join the list into a string.
Time complexity $O(n)$, space complexity $O(n)$, where $n$ is the length of the string.
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| class Solution:
def reverseWords(self, s: str) -> str:
ans = []
i, n = 0, len(s)
while i < n:
while i < n and s[i] == ' ':
i += 1
if i < n:
j = i
while j < n and s[j] != ' ':
j += 1
ans.append(s[i:j])
i = j
return ' '.join(ans[::-1])
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| class Solution {
public String reverseWords(String s) {
List<String> words = new ArrayList<>();
int n = s.length();
for (int i = 0; i < n;) {
while (i < n && s.charAt(i) == ' ') {
++i;
}
if (i < n) {
StringBuilder t = new StringBuilder();
int j = i;
while (j < n && s.charAt(j) != ' ') {
t.append(s.charAt(j++));
}
words.add(t.toString());
i = j;
}
}
Collections.reverse(words);
return String.join(" ", words);
}
}
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