771. Jewels and Stones
771. Jewels and Stones
You're given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.
The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".
给定字符串J 代表石头中宝石的类型，和字符串 S代表你拥有的石头。 S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。
J 中的字母不重复，J 和 S中的所有字符都是字母。字母区分大小写，因此"a"和"A"是不同类型的石头。
示例 1:
输入: J = "aA", S = "aAAbbbb" 输出: 3
示例 2:
输入: J = "z", S = "ZZ" 输出: 0
注意: S 和 J 最多含有50个字母。 J 中的字符不重复。
题解
1.思路：直接两层循环，外层遍历S，内层遍历J，相等直接内层break出去
public int NumJewelsInStones(string J, string S)
{
int jewelCount = 0;
for(int i = 0; i < S.Length; i++)
{
for(int j = 0; j < J.Length; j++)
{
if(J[j] == S[i])
{
jewelCount++;
break;
}
}
}
return jewelCount;
}时间复杂度n*n 空间复杂度0 2.将J存入List，遍历S，看J的List中是否有
List<char> list = new List<char>();
for(int i=0;i<J.Length; i++)
{
list.Add(J[i]);
}
for(int n = 0; n < S.Length; n++)
{
if (J.Contains(S[n]))
{
jewelCount++;
}
}时间复杂度n 空间复杂度n 3.Dictionary的Key存每一个S的char，value存char出现的次数。之后按J作为Key在dic中查找，找到将value加在jewelCount上。
Dictionary<char, int> dic = new Dictionary<char, int>();
for(int i=0;i<S.Length; i++)
{
int temp;
if(dic.TryGetValue(S[i],out temp))
{
dic[S[i]]++;
}
else
{
dic.Add(S[i], 1);
}
}
for(int n = 0; n < J.Length; n++)
{
int temp;
if(dic.TryGetValue(J[n],out temp))
{
jewelCount += temp;
}
}时间复杂度n 空间复杂度n