Leetcode-541. Reverse String II 算法

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。这次比赛略无语，没想到前3题都可以用暴力解。
博客链接：mcf171的博客

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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]

这个题目挺简单的，但是一开始有一个地方写的有一点问题，导致错了一次。

public class Solution { public String reverseStr(String s, int k) { StringBuilder sb = new StringBuilder(""); int start = 0; while(start <= s.length()){ int i = 0,j = 0; if(s.length() - start <= k){ i = s.length() - 1; j = s.length(); }else{ i = start + k - 1; j = start + k; } for( ; i >= start; i--) sb.append(s.charAt(i)); for(; j < s.length() && j < start + 2*k; j ++)sb.append(s.charAt(j)); start += 2*k; }return sb.toString(); } }

【Leetcode-541. Reverse String II】