541. Reverse String II-优快云博客

本文介绍了一种特殊的字符串反转算法，该算法每隔2k个字符反转前k个字符，并针对不同情况进行了详细讨论。提供了完整的Java代码实现，适用于字符串长度和k值在1到10000之间的场景。

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Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less 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 left the other as original.

Example:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Restrictions:

The string consists of lower English letters only.
Length of the given string and k will in the range [1, 10000]

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Solution:

Tips:

Just reverse the code by group, and deal the last words less than k specifically.

Java Code:

public class Solution {
    public String reverseStr(String s, int k) {
        if (s == null || k < 2) {
            return s;
        }
        
        StringBuilder result = new StringBuilder();
        int i = 0;
        int kCount = 0;
        
        while (i < s.length()) {
            int n = k;
            int beginIdx = kCount % 2 == 0 ? i + k - 1 : i;
            int step = kCount % 2 == 0 ? -1 : 1;
            if (i + k > s.length()) {
                break;
            }
            kCount++;
            i += k;
            while (n-- > 0) {
                result.append(s.charAt(beginIdx));
                beginIdx += step;
            }
        }
        StringBuilder tailStr = new StringBuilder(s.substring(i, s.length()));
        if (kCount % 2 == 0) {
            result.append(tailStr.reverse());
        } else {
            result.append(tailStr);
        }
        
        return new String(result);
    }
}