LeetCode 271. Encode and Decode Strings

ActiveCoder

于 2016-05-24 12:41:54 发布

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分类专栏： LeetCode 题解

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本文介绍了一种字符串编码和解码的方法。编码通过在每个字符串前添加长度和特殊符号来实现，便于后续的解码过程。解码则依据这些特殊符号及长度信息将编码后的字符串还原为原始形式。示例代码展示了如何对一组字符串进行编码和解码。

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This is to encode the string not to COMPRESS the string.

This is one possible answer.

#include <string>
#include <vector>
#include <iostream>
using namespace std;

string encode(vector<string>& strs) {
  string res = "";
  for(int i = 0; i < strs.size(); ++i) {
    res += to_string(strs[i].size()) + "#" + strs[i];
  }
  return res;
}

vector<string> decode(string s) {
  vector<string> res;
  int i = 0, j = 0;
  while(i < s.size()) {
    if(s[i] != '#') {i++; continue;}
    else {
      int len = stoi(s.substr(j, i - j));
      res.push_back(s.substr(i + 1, len));
      i = i + 1 + len;
      j = i;
    }
  }
  //while(s.size() > 0) {
  //  int pos = s.find_first_of('#');   // if not know this function, better to use two pointers.
  //  int len = stoi(s.substr(0, pos));
  //  res.push_back(s.substr(pos + 1, len));
  //  s = s.substr(pos + len + 1);
  //}
  return res;
}

int main(void) {
  vector<string> words{"tammy", "is", "a", "good", "girl"};
  string s = encode(words);
  cout << endl;
  vector<string> res = decode(s);
  for(int i = 0; i < res.size(); ++i) cout << res[i] << endl;
}