将n进制的数组压缩成字符串（0-9 a-z）同时解压

例如一个3进制的数组： [1 1 2 2 2 0 0] 用一个字符串表示。。。

此类题目要明确两点：

1. 打表：用数组下标索引字符，同时注意如果从字符对应回数字： 

int index = (str[i] >= '0' && str[i] <= '9') ? (str[i]-'0'-radix):(str[i]-'a'-radix + 10);

2. 注意低位在前还是高位在前，如果先来的是 低位*radix^i 即可。

3. 统计每几个radix进制数组成一位，利用bits来表示。。。

这破题主要是麻烦。。。

#include <assert.h>
#include <algorithm>
#include <vector>
using namespace std;
const int radix = 3;
string base = "";
string compress(const vector<int>& arr) {
  string res = "";
  int i, j, size = arr.size(), left, bits;
  vector<int> base;
  for (i = 1, j = 0; i*radix+radix < 36; i *= radix, j++); // 剩余部分保持不变，直接追加到结尾
  bits = j;
  left = size - size / bits * bits;
  size -= left;
  for (char ch = '0'; ch <= '9'; ++ch)
    base.push_back(ch);
  for (char ch = 'a'; ch <= 'z'; ++ch)
    base.push_back(ch);
  for (i = 0; i < size; i += bits) {
    int tmp = 0, t = 1;

    for (j = 0; j < bits; ++j) {
      tmp += arr[i+j]*t;
      t *= radix;
    }
    res += base[radix + tmp];
  }
  for (j = 0; j < left; ++j)
    res += base[arr[i+j]];
  return res;
}
vector<int> depress(const string& str) {
  vector<int> res;
  int i, j, len = str.length(), idx, bits;
  for (i = 1, j = 0; i*radix+radix < 36; i *= radix, j++); // 剩余部分保持不变，直接追加到结尾
  bits = j;
  for (i = 0; i < len; ++i) {
    idx = str[i] >= 'a' && str[i] <= 'z' ? (str[i] - 'a' + 10) : (str[i] - '0');
    if (idx < radix) {
      res.push_back(idx);
    }
    else {
      idx -= radix;
      for (j = 0; j < bits; ++j) {
        res.push_back(idx%radix);
        idx /= radix;
      }
    }
  }
  return res;
}
int main() {
  int arr[] = {0,1,2,2,2,2,1,1,1,2,0,0,0,0,0,1};
  vector<int> vec(arr, arr+sizeof(arr)/sizeof(int));
  string str = compress(vec);
  vector<int> res = depress(str);
  return 0;
}

下面的代码多了很多无用的控制字符，不知道为啥。。。

#include <assert.h>
#include <algorithm>
#include <vector>
using namespace std;
const int radix = 4;
string base = "";
string compress(const vector<int>& arr) {
  string res = "";
  
  int i, j, size = arr.size(), left, bits;
  
  for (i = 1, j = 0; i*radix + radix < 36; i *= radix, ++j);
  bits = j;
  left = size - size / bits * bits;
  size = size / bits * bits;


  for (char ch = '0'; ch <= '9'; ++ch)
    base.push_back(ch);
  for (char ch = 'a'; ch <= 'z'; ++ch)
    base.push_back(ch);
  for (i = 0; i < size; i += bits) {
    int index = 0, t = 1;
    for (j = 0; j < bits; ++j) {
      index = index + arr[i+j]*t;
      t *= radix;
    }           
    res.push_back(base[radix+index]);
  }
  for (i = 0; i < left; ++i)
    res.push_back(arr[size+i]+'0');
  return res; 
}


vector<int> depress(const string& str) {
  int len = str.length(), i = 0, j, bits;
  for (i = 1, j = 0; i*radix + radix< 36; i *= radix, ++j);
  bits = j;
  vector<int> res;
  for (i = 0; i < len; ++i) {
    if (str[i]-'0' >= 0 && str[i]-'0' < radix) 
      res.push_back(str[i]-'0');
    else {
      int index = (str[i] >= '0' && str[i] <= '9') ? (str[i]-'0'-radix):(str[i]-'a'-radix + 10);      
      for (j = 0; j < bits; ++j) {
        res.push_back(index%radix);
        index = index/radix;
      }
    }
  }
  return res;
} 
int main() {


  int arr[] = {0,1,2,2,2,2,1,1,1,2,0,0,0,0,0,1};
  vector<int> vec(arr, arr+sizeof(arr) / sizeof(int));




  string str = compress(vec);
  vector<int> res = depress(str);


  return 0;
}