README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0345.Reverse%20Vowels%20of%20a%20String/README_EN.md	Two Pointers String

345. Reverse Vowels of a String

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Description

Given a string s, reverse only all the vowels in the string and return it.

The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and upper cases, more than once.

 

Example 1:

Input: s = "IceCreAm"

Output: "AceCreIm"

Explanation:

The vowels in s are ['I', 'e', 'e', 'A']. On reversing the vowels, s becomes "AceCreIm".

Example 2:

Input: s = "leetcode"

Output: "leotcede"

 

Constraints:

• 1 <= s.length <= 3 * 105
• s consist of printable ASCII characters.

Solutions

Solution 1: Two Pointers

We can use two pointers $i$ and $j$, initially pointing to the start and end of the string respectively.

In each loop, we check whether the character at $i$ is a vowel. If it's not, we move $i$ forward. Similarly, we check whether the character at $j$ is a vowel. If it's not, we move $j$ backward. If $i < j$ at this point, then both characters at $i$ and $j$ are vowels, so we swap these two characters. Then, we move $i$ forward and $j$ backward. We continue the above operations until $i \ge j$.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the size of the character set.

Python3

class Solution:
    def reverseVowels(self, s: str) -> str:
        vowels = "aeiouAEIOU"
        i, j = 0, len(s) - 1
        cs = list(s)
        while i < j:
            while i < j and cs[i] not in vowels:
                i += 1
            while i < j and cs[j] not in vowels:
                j -= 1
            if i < j:
                cs[i], cs[j] = cs[j], cs[i]
                i, j = i + 1, j - 1
        return "".join(cs)

Java

class Solution {
    public String reverseVowels(String s) {
        boolean[] vowels = new boolean[128];
        for (char c : "aeiouAEIOU".toCharArray()) {
            vowels[c] = true;
        }
        char[] cs = s.toCharArray();
        int i = 0, j = cs.length - 1;
        while (i < j) {
            while (i < j && !vowels[cs[i]]) {
                ++i;
            }
            while (i < j && !vowels[cs[j]]) {
                --j;
            }
            if (i < j) {
                char t = cs[i];
                cs[i] = cs[j];
                cs[j] = t;
                ++i;
                --j;
            }
        }
        return String.valueOf(cs);
    }
}

C++

class Solution {
public:
    string reverseVowels(string s) {
        bool vowels[128];
        memset(vowels, false, sizeof(vowels));
        for (char c : "aeiouAEIOU") {
            vowels[c] = true;
        }
        int i = 0, j = s.size() - 1;
        while (i < j) {
            while (i < j && !vowels[s[i]]) {
                ++i;
            }
            while (i < j && !vowels[s[j]]) {
                --j;
            }
            if (i < j) {
                swap(s[i++], s[j--]);
            }
        }
        return s;
    }
};

Go

func reverseVowels(s string) string {
	vowels := [128]bool{}
	for _, c := range "aeiouAEIOU" {
		vowels[c] = true
	}
	cs := []byte(s)
	i, j := 0, len(cs)-1
	for i < j {
		for i < j && !vowels[cs[i]] {
			i++
		}
		for i < j && !vowels[cs[j]] {
			j--
		}
		if i < j {
			cs[i], cs[j] = cs[j], cs[i]
			i, j = i+1, j-1
		}
	}
	return string(cs)
}

TypeScript

function reverseVowels(s: string): string {
    const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
    const cs = s.split('');
    for (let i = 0, j = cs.length - 1; i < j; ++i, --j) {
        while (i < j && !vowels.has(cs[i].toLowerCase())) {
            ++i;
        }
        while (i < j && !vowels.has(cs[j].toLowerCase())) {
            --j;
        }
        [cs[i], cs[j]] = [cs[j], cs[i]];
    }
    return cs.join('');
}

Rust

impl Solution {
    pub fn reverse_vowels(s: String) -> String {
        let vowel = String::from("aeiouAEIOU");
        let mut data: Vec<char> = s.chars().collect();
        let (mut i, mut j) = (0, s.len() - 1);
        while i < j {
            while i < j && !vowel.contains(data[i]) {
                i += 1;
            }
            while i < j && !vowel.contains(data[j]) {
                j -= 1;
            }
            if i < j {
                data.swap(i, j);
                i += 1;
                j -= 1;
            }
        }
        data.iter().collect()
    }
}