1

我有一个输入字符串，需要遍历它并查看它是否与某些单词匹配。我有多个字符串数组，但不确定检查所有数组的字符串的有效方法是什么。

字符串数组：

 string checkPlayType(string printDescription)
{
    const string DeepPassRight[3] = {"deep" , "pass" , "right"};
    const string DeepPassLeft[3] = {"deep" , "pass" , "left"};
    const string DeepPassMiddle[3] = {"deep" , "pass" , "middle"};

    const string ShortPassRight[3] = {"short" , "pass" , "right"};
    const string ShortPassLeft[3] = {"short" , "pass" , "left"};
    const string ShortPassMiddle[3] = {"short" , "pass" , "middle"};

    //Must contain right but not pass
    const string RunRight = "right";
    //Must contain right but not pass
    const string RunLeft = "left";
    //Must contain middle but not pass      
    const string RunMiddle = "middle";

    const string FieldGoalAttempt[2] = {"field" , "goal" };
    const string Punt = "punt";

}

Sample Input: (13:55) (Shotgun) P.Manning pass incomplete short right to M.Harrison.

Assuming this is our only input...
Sample Output: 
Deep Pass Right: 0%
Deep Pass Left: 0%
Deep Pass Middle: 0%
Short Pass Right: 100%
Shor Pass Left:0%
...
..
..
4

3 回答 3

3

你可能想要类似的东西：

void checkPlayType(const std::vector<std::string>& input)
{
    std::set<std::string> s;

    for (const auto& word : input) {
        s.insert(word);
    }
    const bool deep_present = s.count("deep");
    const bool pass_present = s.count("pass");
    const bool right_present = s.count("right");
    const bool left_present = s.count("left");
    // ...

    if (deep_present && pass_present && right_present) { /* increase DeepPassRight counter */}
    if (deep_present && pass_present && left_present) { /* increase DeepPassLeft counter */}
    // ...
}
于 2013-09-09T16:47:30.643 回答
0

尝试正则表达式：

if found "pass" then
    if regexp "(deep|short).*(left|right|middle)"
        Hooray!
    else if regexp "(left|right|middle).*(deep|short)"
        Hooray!
    else
        Aye, Caramba!
else
    Aye, Caramba!
于 2013-09-09T16:28:50.880 回答
0

您可以查看数组并搜索存储在输入字符串中的数组中的单词。使用 std 函数以获得更好的性能。例如：

const string DeepPassRight[3] = {"deep" , "pass" , "right"};
    int i = 0;
    for(;i<3;i++)
    {
        string s = " ";
        s.append(DeepPassRight[i]);
        s.append(" ");
        std::size_t found = printDescription.find(s);
        if (found ==std::string::npos)
           break;

    }
    if(i == 3)
        // printDescription contains all DeepPassRight's members!
if(i== 2)
// just two words were found
于 2013-09-09T16:02:47.217 回答