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: Two Pointers
We can use two pointers i and j to find each word, add it to the result list, then reverse the result list, and finally concatenate it into a string.
The time complexity is O(n), and the space complexity is O(n). Where n is the length of the string.
PythonJavaC++GoTypeScriptRustC#
class Solution:
def reverseWords(self, s: str) -> str:
words = []
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
words.append(s[i:j])
i = j
return " ".join(words[::-1])(code-box)
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);
}
}(code-box)
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;
}
};(code-box)
func reverseWords(s string) string {
words := []string{}
i, n := 0, len(s)
for i < n {
for i < n && s[i] == ' ' {
i++
}
if i < n {
j := i
t := []byte{}
for j < n && s[j] != ' ' {
t = append(t, s[j])
j++
}
words = append(words, string(t))
i = j
}
}
for i, j := 0, len(words)-1; i < j; i, j = i+1, j-1 {
words[i], words[j] = words[j], words[i]
}
return strings.Join(words, " ")
}(code-box)
function reverseWords(s: string): string {
const words: string[] = [];
const n = s.length;
let i = 0;
while (i < n) {
while (i < n && s[i] === ' ') {
i++;
}
if (i < n) {
let j = i;
while (j < n && s[j] !== ' ') {
j++;
}
words.push(s.slice(i, j));
i = j;
}
}
return words.reverse().join(' ');
}(code-box)
impl Solution {
pub fn reverse_words(s: String) -> String {
let mut words = Vec::new();
let s: Vec<char> = s.chars().collect();
let mut i = 0;
let n = s.len();
while i < n {
while i < n && s[i] == ' ' {
i += 1;
}
if i < n {
let mut j = i;
while j < n && s[j] != ' ' {
j += 1;
}
words.push(s[i..j].iter().collect::<String>());
i = j;
}
}
words.reverse();
words.join(" ")
}
}(code-box)
public class Solution {
public string ReverseWords(string s) {
List<string> words = new List<string>();
int n = s.Length;
for (int i = 0; i < n;) {
while (i < n && s[i] == ' ') {
++i;
}
if (i < n) {
System.Text.StringBuilder t = new System.Text.StringBuilder();
int j = i;
while (j < n && s[j] != ' ') {
t.Append(s[j++]);
}
words.Add(t.ToString());
i = j;
}
}
words.Reverse();
return string.Join(" ", words);
}
}(code-box)
Solution 2: String Split
We can use the built-in string split function to split the string into a list of words by spaces, then reverse the list, and finally concatenate it into a string.
The time complexity is O(n), and the space complexity is O(n). Where n is the length of the string.
PythonJavaGoTypeScriptRust
class Solution:
def reverseWords(self, s: str) -> str:
return " ".join(reversed(s.split()))(code-box)
class Solution {
public String reverseWords(String s) {
List<String> words = Arrays.asList(s.trim().split("\\s+"));
Collections.reverse(words);
return String.join(" ", words);
}
}(code-box)
func reverseWords(s string) string {
words := strings.Fields(s)
for i, j := 0, len(words)-1; i < j; i, j = i+1, j-1 {
words[i], words[j] = words[j], words[i]
}
return strings.Join(words, " ")
}(code-box)
function reverseWords(s: string): string {
return s.trim().split(/\s+/).reverse().join(' ');
}(code-box)
impl Solution {
pub fn reverse_words(s: String) -> String {
s.split_whitespace().rev().collect::<Vec<&str>>().join(" ")
}
}(code-box)
