LintCode 3208：String splitting字符串处理入门题

3208 · String splitting
C++
Easy

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Description
This question requires you to complete the function splitString in the file Solution.cpp.

This function takes two arguments.

A string s as the original string to be split.
A non-empty string t is used as a splitting basis.
All the split substrings in order are formed into a vector and returned using the return statement.

The evaluation machine runs main.cpp to call the splitString function in Soluction.cpp by importing a custom function library, and parses the contents of the returned string vector.

There is no need to use the cin statement or the cout statement for this problem.
The input data is stored in the arguments of the function, and the output data is returned using the return statement.
Example
Input Sample 1:

asdf,Hello,ok
,
Output sample 1:

[“asdf”, “Hello”, “ok”]
All commas , of the original string are removed and the string is separated from the comma position.

Input Sample 2:

,
Output sample 2:

[]
The original string is empty, so an empty vector container is returned.

解法1：

/**
 * @param s: A string
 * @param t: A string
 * @return: A vector of sub strings
 */
vector<string> splitString(string &s, string &t) {
    vector<string> res;
    string s2(s);
    while (s2.size() > 0) {
        int pos = s2.find(t);
        if (pos == -1) {
            res.push_back(s2);
            break;
        }
        res.push_back(s2.substr(0, pos));
        s2 = s2.substr(pos + 1);
    }
    return res;
}

解法2：

/**
 * @param s: A string
 * @param t: A string
 * @return: A vector of sub strings
 */
vector<string> splitString(string &s, string &t) {
    vector<string> res;
    int start = 0, pos = 0;
    pos = s.find(t);
    while (pos != -1) {
        res.push_back(s.substr(start, pos - start));
        start = pos + 1;
        pos = s.find(t, start);
        if (pos == -1) res.push_back(s.substr(start));  
    }
    //if (pos == -1) res.push_back(s.substr(pos + 1));
    return res;
}