1.题目描述 Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only
如何将字符串中的10进制数和16进制数提取出来,看以下代码: #include <stdio.h> typedef char TUINT8 ; typedef int TUINT32; TUINT32 Read_DecNumber(const TUINT8* str); TUINT32 Read_HexNumber(const TUINT8* str); int main(void) { int ret = Read_DecNumber("1000"); int d = R
#include <stdio.h> #include <string.h> #define NR(x) sizeof(x)/sizeof(x[0]) int Del_char(const char *input, char *output) { int i , j , k = 0; int flag ; int len ; //判断输入输出的字符串数组是否为空 if(input == NULL || output == NULL) { //如果为空,返回错误码 return -1
String regEx="[\n`~!@#$%^&*()+=|{}':;',\\[\\].<>/?~!@#¥%……&*()——+|{}[]‘::”“’., .?]"; //可以在中括号内加上任何想要替换的字符,实际上是一个正则表达式 String aa = " ";//这里是将特殊字符换为aa字符串," "代表直接去掉 Pattern p = Pattern.compile(regEx); Matcher m = p
点击查看代码 #include<iostream> using namespace std; string str; int n; int main() { cin >> n; while (n --) { cin >> str; int cnt = 0; char c; for (int i = 0; i < str.size(); i++) { int j = i; while (j < str.size() && str[i] == s
[1]a=[8,13,11,6,26,19,24]1)请输出列表a中的奇数项2)请输出列表a中的奇数 解:1) a=[8,13,11,6,26,19,24] print a[::2] Result:>>>[8, 11, 26, 24] 2) a = [8,13,11,6,26,19,24] b = [] for item in a: if item%2 !=0: b.append(item) else: continue print b Result:>>>[13, 1
# -*- coding:utf-8 -*- import sys,os txta = open('a.txt','r') str = '' for line in txta: str += line.strip().decode('utf-8') txta.close() for word in str: print word.encode('utf-8') 直接输出,是会乱码的,得先解码,再编码. 参考网址:http://blog.csdn.net/devil_2009/article/de