public void MimicJoin()
{///本来打算用泛型做这个，但是实在搞不定运算符的问题，这里留个影子，期待飞哥解惑(☆▽☆)
Assert.AreEqual("w!p!z", DataTimeCount.MimicJoin("!", new string[] { "w", "p", "z" }));
Assert.AreEqual("a-b-c-d", DataTimeCount.MimicJoin("-", new string[] { "a", "b", "c", "d" }));
Assert.AreNotEqual(1 - 2 - 3 - 4, DataTimeCount.MimicJoin("-", new string[] { "1", "2", "3", "4" }));
Assert.AreEqual("a@b@c@d", DataTimeCount.MimicJoin("@", new string[] { "a", "b", "c", "d" }));
Assert.AreEqual("a$4/b$4/c$4/d", DataTimeCount.MimicJoin("$4/", new string[] { "a", "b", "c", "d" }));
Assert.AreEqual("aabacad", DataTimeCount.MimicJoin("a", new string[] { "a", "b", "c", "d" }));
}
public void GetCount()
{
//实现GetCount(string container, string target)方法，可以统计出container中有多少个target
//固化要求，已经查过的字符不能在查。不存在重叠【主要是因为重叠想不出来】
string text = "天今天，是昨天的明天，是明天的前一天",
find = "天", find1 = "是", find2 = "you", find3 = "明天", find4 = ",";
Assert.AreEqual(6, DataTimeCount.GetCount(text, find));
Assert.AreEqual(2, DataTimeCount.GetCount(text, find1));
Assert.AreEqual(0, DataTimeCount.GetCount(text, find2));
Assert.AreEqual(2, DataTimeCount.GetCount(text, find3));
Assert.AreEqual(0, DataTimeCount.GetCount(text, find4));
}
