LeetCode原题:Reverse String II
Description: 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]
题目的意思很简单：给定一个字符串S和一个整数k，每2k段反转其前k个字符，剩余的字符串如果长度达不到k，则全部反转，若达到k却小于2k，则只反转前k个字符。
注意到Java中String并没有现成反转字符串的方法，所以我们只能自己写这个反转函数。期间，我们会用到substring(startIndex,endIndex)这个方法，这里一定要注意，startIndex确实是指在字符串中的起始位置，但是endIndex却不是在字符串中的末尾位置，endIndex-1才是。
附Java代码如下：
class Solution { public String reverseStr(String s, int k) { int lenth = s.length(); int parts = lenth/(2*k);//以2k为长度分段 StringBuffer stringRes = new StringBuffer(); for(int i = 0; i < parts; i ++){ //反转前k个字符，并添加 stringRes.append(reverse(s.substring(i*2*k, i*2*k+k))); stringRes.append(s.substring(i*2*k+k,i*2*k+2*k)); } //反转最后一段 if(parts*2*k+k <= lenth){ stringRes.append(reverse(s.substring(parts*2*k, parts*2*k+k))); stringRes.append(s.substring(parts*2*k+k,lenth)); }else{ stringRes.append(reverse(s.substring(parts*2*k, lenth))); } return stringRes.toString(); } public String reverse(String s){ //返回此字符串的反转 StringBuffer strBuffer = new StringBuffer(); char temp; for(int i = 0; i < s.length(); i ++){ temp = s.charAt(s.length()-i-1); strBuffer.append(temp); } return strBuffer.toString(); } }