[字符串与数组] Q:Write a method to replace all spaces in a string with ‘%20’ 题目:写一个算法将一个字符串中的空格替换成%20 解答: 很直观的解法,首先统计出字符串中空格个数,然后分配新的内存空间,依次从头到尾复制原字符串到新字符串中,遇到空格,则复制%20这三个字符.还有没有其他更好点的方法呢?? char* replace(char* str){ int len=strlen(str); int spaceNum=0; int…

使用“;”替换过字符串中的“,” public class Test01 {public static void main(String[] args) {String number = "123,456,5234,52345,63456,7456,7";String newNumber = number.replace(",", ";");System.out.println(newNumber);} } 结果: 123;456;5234;52…

JS正则表达式获取字符串中得特定字符,通过replace的回调函数获取. 实现的效果:在字符串中abcdefgname='test'sddfhskshjsfsjdfps中获取name的值test  实现的机制:通过replace的回调函数获取.  代码:  var str = "abcdefgname='test'sddfhskshjsfsjdfps";  var reg = /name='((\w|-|\s)+)/ig;  str.replace(reg, function() { …

/// <summary> /// 计算字符串中子串出现的次数 /// </summary> /// <param name="str">字符串</param> /// <param name="substring">子串</param> /// <returns>出现的次数</returns> private int SubstringCount(string str,…

[字符串与数组] Q:Assume you have a method isSubstring which checks if one word is a substring of another Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using only one call to isSubstring (i e , “waterbottle” is a rotation of “e…

[链表] Q:Write code to remove duplicates from an unsorted linked list      FOLLOW UP      How would you solve this problem if a temporary buffer is not allowed? 题目:编码实现从无序链表中移除重复项.          如果不能使用临时缓存,你怎么编码实现? 解答: 方法一:不使用额外的存储空间,直接在原始链表上进行操作.首先用一个指针指向链…

','**') --下面是结果集 /* ----------- 12345678** */ SELECT replace(CONVERT(varchar(),GETDATE(),),'-','') --下面是结果集 /* ------------ 20140325 */…

update CfmRcd set reconsource=replace(reconsource,'''',''), cmffile =replace(cmffile,'''',''), cfmdate=replace(cfmdate,'''',''), '20170721'…

替换字符串中的子串 任务: 给定一个字符串,通过查询一个字符串替换字典,将字符串中被标记的子字符串替换掉. 解决方案: >>> import string >>> new_style = string.Template('this is $thing') #给substitute 方法传入一个字典参数并调用 >>> print new_style.substitute({'thing':5}) this is 5 >>> print…

大家都知道字符串在python中是不可变数据类型,那么我们如何替换字符串中指定位置的字符呢? 字符串转换列表替换并转换解决: def replace_char(string,char,index): string = list(string) string[index] = char return ''.join(string)…