原題
https://www.codewars.com/kata/reverse-or-rotate/train/cpp
題目
The input is a string
strof digits. Cut the string into chunks (a chunk here is a substring of the initial string) of sizesz(ignore the last chunk if its size is less thansz).
If a chunk represents an integer such as the sum of the cubes of its digits is divisible by2, reverse that chunk; otherwise rotate it to the left by one position. Put together these modified chunks and return the result as a string.
If
szis<= 0or if str is empty return""
szis greater (>) than the length ofstrit is impossible to take a chunk of sizeszhence return"".
Examples:
revrot("123456987654", 6) --> "234561876549"
revrot("123456987653", 6) --> "234561356789"
revrot("66443875", 4) --> "44668753"
revrot("66443875", 8) --> "64438756"
revrot("664438769", 8) --> "67834466"
revrot("123456779", 8) --> "23456771"
revrot("", 8) --> ""
revrot("123456779", 0) --> ""
revrot("563000655734469485", 4) --> "0365065073456944"
分析
本題主要是數(shù)字字符串處理。主要分為三個要點。
- 數(shù)組字符串累計求和
- 字符串反轉(zhuǎn)
- 字符串旋轉(zhuǎn)
這三個問題都可以使用STL算法函數(shù)模板解決车猬。
參考答案
#include <string>
#include <numeric>
using namespace std;
class RevRot
{
public:
static std::string revRot(const std::string &str, unsigned int sz){
if(0 == sz or str.empty() or sz > str.size()){
return "";
}
size_t count(str.size()/sz);
string res(str,0,count*sz);// 保留滿足sz部分
auto it = res.begin();
while(it < res.end()){
if(accumulate(it,it+sz,'0',[](char r,char l){return (r-'0')+(l-'0');})%2){
rotate(it,it+1,it+sz);
}else{
reverse(it,it+sz);
}
it += sz;
}
return res;
};
};
說明
無
其它
無
