LeetCode in Kotlin

541. Reverse String II

Easy

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.

Example 1:

Input: s = “abcdefg”, k = 2

Output: “bacdfeg”

Example 2:

Input: s = “abcd”, k = 2

Output: “bacd”

Constraints:

• 1 <= s.length <= 104
• s consists of only lowercase English letters.
• 1 <= k <= 104

Solution

class Solution {
    fun reverseStr(s: String, k: Int): String {
        val res = StringBuilder()
        var p1 = 0
        var p2 = 2 * k - 1
        if (s.length < k) {
            res.append(reverse(s))
            return res.toString()
        }
        while (p1 < s.length) {
            if (s.length < p1 + k) {
                res.append(reverse(s.substring(p1)))
            } else {
                res.append(reverse(s.substring(p1, p1 + k)))
                if (s.length < p2 + 1) {
                    res.append(s.substring(p1 + k))
                } else {
                    res.append(s, p1 + k, p2 + 1)
                }
            }
            p1 = p1 + 2 * k
            p2 = p2 + 2 * k
        }
        return res.toString()
    }

    private fun reverse(s: String): String {
        val sb = StringBuilder()
        sb.append(s)
        sb.reverse()
        return sb.toString()
    }
}

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