摘要： 题目：The count-and-say sequence is the sequence of integers beginning as follows:1, 11, 21, 1211, 111221, ...1is read off as"one 1"or11.11is read off as"two 1s"or21.21is read off as"one 2, thenone 1"or1211.Given an integern, generate thenthsequence.Note: The sequence of i 阅读全文

摘要： 题目：Given a roman numeral, convert it to an integer.Input is guaranteed to be within the range from 1 to 3999.解题思路：我们首先来明确一下罗马数字与阿拉伯数字的换算规则：如果当前数字比前一个大，说明这一段的值应该是当前这个值减去上一个值，比如IV = 5 – 1；否则，将当前值加入到结果中，然后开始下一段记录，比如VI = 5 + 1, II=1+1。而罗马数字与阿拉伯数字对应变换是：I对应1，V对应5，X对应10，L对应50，C对应100，D对应500，M对应1000。因此，只需要从前 阅读全文

摘要： 题目：Write a function to find the longest common prefix string amongst an array of strings.解题思路：先找出最短长度的字符串，然后将所有字符串与最短长度的字符串做比较，遇到不同的字符就返回最短长度字符串的前一部分子串。代码：class Solution {public: string longestCommonPrefix(vector &strs) { int MinSize=INT_MAX; int MinIndex=0; if(strs.empty())r... 阅读全文

摘要： 题目：Given two binary strings, return their sum (also a binary string).For example,a ="11"b ="1"Return"100".解题思路1：判断当前字符所表示的数字，产生输出和进位。缺点：程序比较复杂。代码1：class Solution {public: string addBinary(string a, string b) { char carry='0'; string c; if(a.empty())return b; if( 阅读全文

摘要： 题目：Implementatoito convert a string to an integer.Hint:Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.Notes:It is intended for this problem to be specified vaguely (ie, no given input specs). You are re 阅读全文