KamaCoder-55, Right-Rotated String

I Problem

The right rotation operation of a string is to transfer several characters at the end of the string to the front of the string. Given a string s and a positive integer k, please write a function to move the following k characters in the string to the front of the string, achieving the right rotation operation of the string.

For example, for the input string "abcdefg" and integer 2, the function should convert it to "fgabcde".

Example 1
Input: k = 2, s = "abcdefg"
Output: "fgabcde"

Example 2
Input: k = 5, s = "abcdefg"
Output: "cdefgab"

Constraints

• 1 <= k < 10000
• 1 <= s.length < 10000

Related Topics

• string

II Solution

Approach 1: Brute Force

pub fn right_rotate_string(mut s: String, mut k: usize) -> String {
    let p = unsafe { s.as_bytes_mut() };
    let len = p.len();
    if k == len || k == 0 {
        return s;
    }

    if k <= len / 2 {
        // rotate right
        while k != 0 {
            let temp = p[len - 1];
            for i in (0..len - 1).rev() {
                p[i + 1] = p[i];
            }
            p[0] = temp;
            k -= 1;
        }
    } else {
        // rotate left
        k = len - k;
        while k != 0 {
            let temp = p[0];
            for i in 1..len {
                p[i - 1] = p[i];
            }
            p[len - 1] = temp;
            k -= 1;
        }
    }

    s
}

public String rightRotateString(String s, int k) {
    char[] chars = s.toCharArray();
    int len = chars.length;
    if (k == len || k == 0) {
        return s;
    }

    if (k <= len / 2) {
        // rotate right
        while (k-- != 0) {
            char temp = chars[len - 1];
            for (int i = len - 2; i >= 0; i--) {
                chars[i + 1] = chars[i];
            }
            chars[0] = temp;
        }
    } else {
        // rotate left
        k = len - k;
        while (k-- != 0) {
            char temp = chars[0];
            for (int i = 1; i < len; i++) {
                chars[i - 1] = chars[i];
            }
            chars[len - 1] = temp;
        }
    }

    return new String(chars);
}

Approach 2: Optimize Brute Force

pub fn right_rotate_string(mut s: String, mut k: usize) -> String {
    let p = unsafe { s.as_bytes_mut() };
    let len = p.len();
    if k == len || k == 0 {
        return s;
    }

    if k <= len / 2 {
        // rotate right
        let mut temp = vec![0; k];
        for i in 0..k {
            temp[i] = p[len - k + i];
        }
        for i in (0..len - k).rev() {
            p[i + k] = p[i];
        }
        for i in 0..k {
            p[i] = temp[i];
        }
    } else {
        // rotate left
        k = len - k;
        let mut temp = vec![0; k];
        for i in 0..k {
            temp[i] = p[i];
        }
        for i in k..len {
            p[i - k] = p[i];
        }
        for i in 0..k {
            p[len - k + i] = temp[i];
        }
    }

    s
}

public String rightRotateString(String s, int k) {
    char[] chars = s.toCharArray();
    int len = chars.length;
    if (k == len || k == 0) {
        return s;
    }

    if (k <= len / 2) {
        // rotate right
        char[] temp = new char[k];
        for (int i = 0; i < k; i++) {
            temp[i] = chars[len - k + i];
        }
        for (int i = len - 1 - k; i >= 0; i--) {
            chars[i + k] = chars[i];
        }
        for (int i = 0; i < k; i++) {
            chars[i] = temp[i];
        }
    } else {
        // rotate left
        k = len - k;
        char[] temp = new char[k];
        for (int i = 0; i < k; i++) {
            temp[i] = chars[i];
        }
        for (int i = k; i < len; i++) {
            chars[i - k] = chars[i];
        }
        for (int i = 0; i < k; i++) {
            chars[len - k + i] = temp[i];
        }
    }
    
    return new String(chars);
}