917. Reverse Only Letters

Description

Given a string s, reverse the string according to the following rules:

All the characters that are not English letters remain in the same position.
All the English letters (lowercase or uppercase) should be reversed.

Return s after reversing it.

Example 1:

Input: s = "ab-cd"
Output: "dc-ba"

Example 2:

Input: s = "a-bC-dEf-ghIj"
Output: "j-Ih-gfE-dCba"

Example 3:

Input: s = "Test1ng-Leet=code-Q!"
Output: "Qedo1ct-eeLg=ntse-T!"

Constraints:

1 <= s.length <= 100
s consists of characters with ASCII values in the range [33, 122].
s does not contain '\"' or '\\'.

Solutions

Solution 1: Two Pointers

We use two pointers \(i\) and \(j\) to point to the head and tail of the string respectively. When \(i < j\), we continuously move \(i\) and \(j\) until \(i\) points to an English letter and \(j\) points to an English letter, then we swap \(s[i]\) and \(s[j]\). Finally, we return the string.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the length of the string.

Python3JavaC++GoTypeScriptRust

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

class Solution {
    public String reverseOnlyLetters(String s) {
        char[] cs = s.toCharArray();
        int i = 0, j = cs.length - 1;
        while (i < j) {
            while (i < j && !Character.isLetter(cs[i])) {
                ++i;
            }
            while (i < j && !Character.isLetter(cs[j])) {
                --j;
            }
            if (i < j) {
                char t = cs[i];
                cs[i] = cs[j];
                cs[j] = t;
                ++i;
                --j;
            }
        }
        return new String(cs);
    }
}

class Solution {
public:
    string reverseOnlyLetters(string s) {
        int i = 0, j = s.size() - 1;
        while (i < j) {
            while (i < j && !isalpha(s[i])) {
                ++i;
            }
            while (i < j && !isalpha(s[j])) {
                --j;
            }
            if (i < j) {
                swap(s[i++], s[j--]);
            }
        }
        return s;
    }
};

func reverseOnlyLetters(s string) string {
    cs := []rune(s)
    i, j := 0, len(s)-1
    for i < j {
        for i < j && !unicode.IsLetter(cs[i]) {
            i++
        }
        for i < j && !unicode.IsLetter(cs[j]) {
            j--
        }
        if i < j {
            cs[i], cs[j] = cs[j], cs[i]
            i++
            j--
        }
    }
    return string(cs)
}

function reverseOnlyLetters(s: string): string {
    const cs = [...s];
    let [i, j] = [0, cs.length - 1];
    while (i < j) {
        while (!/[a-zA-Z]/.test(cs[i]) && i < j) {
            i++;
        }
        while (!/[a-zA-Z]/.test(cs[j]) && i < j) {
            j--;
        }
        [cs[i], cs[j]] = [cs[j], cs[i]];
        i++;
        j--;
    }
    return cs.join('');
}

impl Solution {
    pub fn reverse_only_letters(s: String) -> String {
        let mut cs: Vec<char> = s.chars().collect();
        let n = cs.len();
        let mut l = 0;
        let mut r = n - 1;
        while l < r {
            if !cs[l].is_ascii_alphabetic() {
                l += 1;
            } else if !cs[r].is_ascii_alphabetic() {
                r -= 1;
            } else {
                cs.swap(l, r);
                l += 1;
                r -= 1;
            }
        }
        cs.iter().collect()
    }
}

917. Reverse Only Letters

Description

Solutions

Solution 1: Two Pointers

Comments