Reverse or rotate? -- 6 kyu

原题

https://www.codewars.com/kata/reverse-or-rotate/train/cpp

题目

The input is a string str of digits. Cut the string into chunks (a chunk here is a substring of the initial string) of size sz (ignore the last chunk if its size is less than sz).

If a chunk represents an integer such as the sum of the cubes of its digits is divisible by 2, reverse that chunk; otherwise rotate it to the left by one position. Put together these modified chunks and return the result as a string.

• sz is <= 0 or if str is empty return “”
• sz is greater (>) than the length of str it is impossible to take a chunk of size sz hence return “”

example:
revrot(“563000655734469485”, 4) –> “0365065073456944”

分析

此题是指输入一个数字字符串和块的长度sz，然后以sz将串划分成多个块，进行遍历

• 如果块每一个数字之和为偶数，则反转该块；
• 否则将以该块第二个为中心点，旋转
• 如果串长小于块长或者块长<=0，都返回一个空string。
• 最后一个块的长度如果不足sz，则忽略

代码

#include <string>class RevRot{public:    static std::string revRot(const std::string &strng, unsigned int sz);};std::string RevRot::revRot(const std::string &strng, unsigned int sz){  size_t len = strng.size();  if(sz == 0 || sz == 0||sz > len){      return std::string();  }  unsigned chunk_num = len/sz;  auto chunk_begin = begin(strng);  auto chunk_end = chunk_begin+sz;  size_t str_len = chunk_num*sz;  std::string str;  str.resize(str_len);  auto str_begin = begin(str);  for(size_t i = 0;i < chunk_num;i++){   unsigned int sum = 0;   for(size_t j = i*sz;j < (i+1)*sz;j++){     sum += (int)(strng[j]-'0');   }   if(sum%2 == 0){      reverse_copy(chunk_begin,chunk_end,str_begin);   } else{      rotate_copy(chunk_begin,chunk_begin+1,chunk_end,str_begin);   }   chunk_begin+=sz;   chunk_end+=sz;   str_begin+=sz;  }  return str;}