标签：== bsp builder rom gre div public put letter
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:
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
标签：== bsp builder rom gre div public put letter
原文地址：http://www.cnblogs.com/wxisme/p/7710740.html
