[LeetCode&Python] Problem 541. Reverse String II
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]
class Solution(object):
def reverseStr(self, s, k):
"""
:type s: str
:type k: int
:rtype: str
"""
n=len(s)
i=0
ans=''
while i<n:
if i+k-1>=n:
ans+=(s[i:])[::-1]
break
elif i+2*k-1>=n:
ans+=(s[i:i+k])[::-1]+s[i+k:]
break
else:
ans+=(s[i:i+k])[::-1]+s[i+k:i+2*k]
i=i+2*k
return ans
[LeetCode&Python] Problem 541. Reverse String II的更多相关文章
- 【leetcode_easy】541. Reverse String II
problem 541. Reverse String II 题意: 给定一个字符串,每隔k个字符翻转这k个字符,剩余的小于k个则全部翻转,否则还是只翻转剩余的前k个字符. solution1: cl ...
- [LeetCode] 344 Reverse String && 541 Reverse String II
原题地址: 344 Reverse String: https://leetcode.com/problems/reverse-string/description/ 541 Reverse Stri ...
- leetcode 344. Reverse String 、541. Reverse String II 、796. Rotate String
344. Reverse String 最基础的旋转字符串 class Solution { public: void reverseString(vector<char>& s) ...
- leadcode 541. Reverse String II
package leadcode; /** * 541. Reverse String II * Easy * 199 * 575 * * * Given a string and an intege ...
- LeetCode 541. Reverse String II (反转字符串 II)
Given a string and an integer k, you need to reverse the first k characters for every 2k characters ...
- 【LeetCode】541. Reverse String II 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 Java解法 Python解法 日期 题目地址:ht ...
- [LeetCode&Python] Problem 557. Reverse Words in a String III
Given a string, you need to reverse the order of characters in each word within a sentence while sti ...
- 541 Reverse String II 反转字符串 II
给定一个字符串和一个整数 k,你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转.如果剩余少于 k 个字符,则将剩余的所有全部反转.如果有小于 2k 但大于或等于 k 个字符,则反转前 ...
- [LeetCode&Python] Problem 917. Reverse Only Letters
Given a string S, return the "reversed" string where all characters that are not a letter ...
随机推荐
- 转:Linux环境变量设置方法总结 PATH、LD_LIBRARY_PATH
转: https://www.linuxidc.com/Linux/2017-03/142338.htm 文章写比较全 转载记录 Linux环境变量设置方法总结 PATH.LD_LIBRARY_P ...
- ie edge 自动给数字加下划线
<meta name="format-detection" content="telephone=no,email=no,address=no">
- ubuntu ssh前后台切换命令相关
后台运行:命令+& 例如 sleep 60 & jobs -l 显示job的pid和状态 ps 显示用户进程 将第一个job切换回前台:fg 1 放到后台:bg 1 cltr + z ...
- RBAC功能模块
- break&&continue
break和continue的区别: 1. 当它们用在循环语句的循环体时,break用于立即退出本层循环,而continue仅仅结束本次循环(本次循环体内不执行continue语句后的其它语句,但下一 ...
- oracle servicename 与SID的区别
http://blog.csdn.net/z69183787/article/details/25706269
- 一个canvas的demo
该demo放于tomcat下运行,否则出现跨域错误 <!DOCTYPE html> <html> <head> <meta charset="utf ...
- python全栈开发笔记---基本数据类型--字符串魔法
字符串: def capitalize(self, *args, **kwargs) test = "aLxs" v = test.capitalize() #capitalize ...
- confirm消息对话框
function rec(){ var mymessage= confirm("你是女孩?") ; if(mymessage==true) { document.write(&qu ...
- DevExpress WPF v18.2新版亮点(一)
买 DevExpress Universal Subscription 免费赠 万元汉化资源包1套! 限量15套!先到先得,送完即止!立即抢购>> 行业领先的.NET界面控件2018年第 ...