移除重复字符很简单,这里是最笨,也是最简单的一种.问题关键是理解排序的意义: # coding=utf-8 #learning at jeapedu in 2013/10/26 #移除给定字符串中重复字符,参数s是字符串 def removeDuplicate(s): s = list(s) s.sort() #对给定字符串排序,不排序可能移除不完整 for i in s: while s.count(i) > 1: s.remove(i) s = "".join(s) #把列表
因为C#的code,感觉实现这个还是比较容易,只是SB.所以,在面试时候,以为不会这么容易,所以,我先试着用了Dictionary去实现,发现有困难.然后改回来用StringBuilder实现,这个因为比较简单了,总以为这样会有问题.所以不知道这样是不是对的. 结果当然很不理想,我准备后面学习一下C++,看看这个题目用C++实现有什么值得注意的地方.还是有什么精华所在. using System; using System.Collections.Generic; using System.Li
If order does not matter, you can use foo = "mppmt" "".join(set(foo)) set() will create a set of unique letters in the string, and "".join() will join the letters back to a string in arbitrary order. If order does matter,
Qualys项目中写到将ServerIP以“,”分割后插入数据库并将重复的IP去除后发送到服务器进行Scan,于是我写了下面的一段用来剔除重复IP: //CR#1796870 modify by v-yangwu, remove the IPs which are repeated. string[] arrayIP = ipAll.Split(','); List<string> listIP = new List<string>(); foreach (string ip in
#include<stdio.h> /*使用n=strlen(s)时加这个#include<string.h>*/ int main(void) { ];/*定义变量*/ int i, j, n, k; printf("请输入一串有重复字符的字符串:\n"); gets_s(s);/*输入字符串*/ ; s[n] != '\0'; n++); /*用于得到字符串长度,也可以使用n=strlen(s);*/ ; i < n ; i++) { ; j <
三种java 去掉字符串中的重复字符函数 public static void main(string[] args) { system.out.println(removerepeatedchar("ddddccffgd")); } public static string removerepeatedchar(string s) { if (s == null) return s; stringbuilder sb = new stringbuilder(); , len = s.
Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times. Example 1: Input: s = "aaabb", k = 3 Output: 3 The longest substring is "aaa"
