1、题目
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]
2、代码实现
public class Solution { public String reverse(String s) { if (s == null || s.length() == 0) { return null; } char[] chars = s.toCharArray(); int length = chars.length; int start = 0, end = length - 1; while (start < end) { char tmp = chars[start]; chars[start++] = chars[end]; chars[end--] = tmp; } String result = ""; for (char c : chars) result += c; return result; } public String reverseStr(String s, int k) { if (null == s || s.length() == 0) return null; int length = s.length(); int one = 1; int count = length % k == 0 ? length / k : length / k + 1; String result = ""; for (int i = 1; i <= count; i++) { if (one % 2 == 1) result += reverse(s.substring(k * (i -1), (k * i > length ? length : k * i))); else result += s.substring(k * (i -1), (k * i > length ? length : k * i)); one++; } return result; } }
3、总结
先把问题化为一小步，分治的思想，比如我们先实现字符串的反转功能，然后再更具条件来实现，哪些子字符串需要反转
