首页 > 其他 > 详细

LeetCode - 541. Reverse String II

时间:2017-10-22 20:47:39      阅读:277      评论:0      收藏:0      [点我收藏+]

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:

  1. The string consists of lower English letters only.
  2. Length of the given string and k will in the range [1, 10000]
class Solution {
    public String reverseStr(String s, int k) {
        StringBuilder ret = new StringBuilder();   
        if (s.length() <= k) {
            for (int i=0; i<s.length(); i++)
                ret.append(s.charAt(s.length()-i-1));
            return ret.toString();
        }
        
        int kc = s.length() / k;
        if ((s.length() % k) != 0) kc ++;
        for (int ki=0; ki<kc; ki++) {
            if ((ki&1)==0)
                for (int i=Math.min((ki+1)*k-1,s.length()-1); i>=ki*k; i--)
                    ret.append(s.charAt(i));
            else
                for (int i=ki*k; i<(ki+1)*k&&i<s.length(); i++)
                    ret.append(s.charAt(i));
        }
        return ret.toString();
    }
}

 

LeetCode - 541. Reverse String II

原文:http://www.cnblogs.com/wxisme/p/7710740.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!