541 Reverse String II

原创于 2018-04-19 15:30:13 发布 · 137 阅读

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本文介绍了一种特殊的字符串反转算法，该算法每隔2k个字符将前k个字符进行反转。通过递归实现，适用于长度在1到10000之间的字符串。

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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]

class Solution {
  public String reverseStr(String s, int k) {
    if (s.length() <= k) { return new StringBuilder(s).reverse().toString(); }
    else if (s.length() < 2 * k) { return new StringBuilder(s.substring(0, k)).reverse().toString() + s.substring(k); }
    return new StringBuilder(s.substring(0, k)).reverse().toString() + s.substring(k, 2* k) + reverseStr(s.substring(2 * k), k);
  }
}