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151. Reverse Words in a String

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 <= 104
  • s 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.

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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])

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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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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 := []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 <