[LeetCode] 271. Encode and Decode Strings 加码解码字符串
Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.
Machine 1 (sender) has the function:
string encode(vector<string> strs) {
// ... your code
return encoded_string;
}
Machine 2 (receiver) has the function:
vector<string> decode(string s) {
//... your code
return strs;
}
So Machine 1 does:
string encoded_string = encode(strs);
and Machine 2 does:
vector<string> strs2 = decode(encoded_string);
strs2 in Machine 2 should be the same as strs in Machine 1.
Implement the encode and decode methods.
Note:
- The string may contain any possible characters out of 256 valid ascii characters. Your algorithm should be generalized enough to work on any possible characters.
- Do not use class member/global/static variables to store states. Your encode and decode algorithms should be stateless.
- Do not rely on any library method such as
evalor serialize methods. You should implement your own encode/decode algorithm.
给字符加码再解码，先有码再无码，题目没有限制加码的方法，那么只要能成功的把有码变成无码就行了，具体变换方法自己设计。
Java:
public String encode(List<String> strs) {
StringBuffer out = new StringBuffer();
for (String s : strs)
out.append(s.replace("#", "##")).append(" # ");
return out.toString();
}
public List<String> decode(String s) {
List strs = new ArrayList();
String[] array = s.split(" # ", -1);
for (int i=0; i<array.length-1; ++i)
strs.add(array[i].replace("##", "#"));
return strs;
}
Java: with streaming
public String encode(List<String> strs) {
return strs.stream()
.map(s -> s.replace("#", "##") + " # ")
.collect(Collectors.joining());
}
public List<String> decode(String s) {
List strs = Stream.of(s.split(" # ", -1))
.map(t -> t.replace("##", "#"))
.collect(Collectors.toList());
strs.remove(strs.size() - 1);
return strs;
}
Java:
// Encodes a list of strings to a single string.
public String encode(List<String> strs) {
StringBuilder output = new StringBuilder();
for(String str : strs){
// 对于每个子串，先把其长度放在前面，用#隔开
output.append(String.valueOf(str.length())+"#");
// 再把子串本身放在后面
output.append(str);
}
return output.toString();
}
// Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> res = new LinkedList<String>();
int start = 0;
while(start < s.length()){
// 找到从start开始的第一个#，这个#前面是长度
int idx = s.indexOf('#', start);
int size = Integer.parseInt(s.substring(start, idx));
// 根据这个长度截取子串
res.add(s.substring(idx + 1, idx + size + 1));
// 更新start为子串后面一个位置
start = idx + size + 1;
}
return res;
}
Java: better
public String encode(List<String> strs) {
StringBuffer result = new StringBuffer();
if(strs == null || strs.size() == 0)
return result.toString();
for(String str: strs){
result.append(str.length());
result.append("#");
result.append(str);
}
return result.toString();
}
// Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> result = new ArrayList();
if(s == null || s.length() == 0)
return result;
int current = 0;
while(true){
if(current == s.length())
break;
StringBuffer sb = new StringBuffer();
while(s.charAt(current) != '#'){
sb.append(s.charAt(current));
current++;
}
int len = Integer.parseInt(sb.toString());
int end = current + 1 + len;
result.add(s.substring(current+1, end));
current = end;
}
return result;
}
Java: Time Complexity - O(n)， Space Complexity - O(1)
public class Codec {
// Encodes a list of strings to a single string.
public String encode(List<String> strs) {
if(strs == null || strs.size() == 0) {
return "";
}
StringBuilder sb = new StringBuilder();
for(String s : strs) {
int len = s.length();
sb.append(len);
sb.append('/');
sb.append(s);
}
return sb.toString();
}
// Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> res = new ArrayList<>();
if(s == null ||s.length() == 0) {
return res;
}
int index = 0;
while(index < s.length()) {
int forwardSlashIndex = s.indexOf('/', index);
int len = Integer.parseInt(s.substring(index, forwardSlashIndex));
res.add(s.substring(forwardSlashIndex + 1, forwardSlashIndex + 1 + len));
index = forwardSlashIndex + 1 + len;
}
return res;
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(strs));
Java: Time Complexity - O(n)， Space Complexity - O(n)
public class Codec {
// Encodes a list of strings to a single string.
public String encode(List<String> strs) {
StringBuilder sb = new StringBuilder();
for (String s : strs) {
sb.append(s.length()).append('#').append(s);
}
return sb.toString();
}
// Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> res = new ArrayList<>();
if (s == null || s.length() == 0) return res;
for (int lo = 0, i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '#') {
int len = Integer.parseInt(s.substring(lo, i));
res.add(s.substring(i + 1, i + 1 + len));
lo = i + 1 + len;
i = i + 1 + len;
}
}
return res;
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(strs));
Python:
# Time: O(n)
# Space: O(1)
class Codec:
def encode(self, strs):
"""Encodes a list of strings to a single string.
:type strs: List[str]
:rtype: str
"""
encoded_str = ""
for s in strs:
encoded_str += "%0*x" % (8, len(s)) + s
return encoded_str
def decode(self, s):
"""Decodes a single string to a list of strings.
:type s: str
:rtype: List[str]
"""
i = 0
strs = []
while i < len(s):
l = int(s[i:i+8], 16)
strs.append(s[i+8:i+8+l])
i += 8+l
return strs
C++:
class Codec {
public:
// Encodes a list of strings to a single string.
string encode(vector<string>& strs) {
string res = "";
for (auto a : strs) {
res.append(to_string(a.size())).append("/").append(a);
}
return res;
}
// Decodes a single string to a list of strings.
vector<string> decode(string s) {
vector<string> res;
int i = 0;
while (i < s.size()) {
auto found = s.find("/", i);
int len = atoi(s.substr(i, found).c_str());
res.push_back(s.substr(found + 1, len));
i = found + len + 1;
}
return res;
}
};
C++:
class Codec {
public:
// Encodes a list of strings to a single string.
string encode(vector<string>& strs) {
string res = "";
for (auto a : strs) {
res.append(to_string(a.size())).append("/").append(a);
}
return res;
}
// Decodes a single string to a list of strings.
vector<string> decode(string s) {
vector<string> res;
while (!s.empty()) {
int found = s.find("/");
int len = atoi(s.substr(0, found).c_str());
s = s.substr(found + 1);
res.push_back(s.substr(0, len));
s = s.substr(len);
}
return res;
}
};
All LeetCode Questions List 题目汇总
