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".
Example 1:
Input:J = "aA", S = "aAAbbbb"Output:3
Example 2:
Input:J = "z", S = "ZZ"Output:0
Note:
S?and?J?will consist of letters and have length at most 50.
The characters in?J?are distinct.
第一次提交代碼
class Solution {
public:
? ? int numJewelsInStones(string J, string S) {
? ? ? ? int li=J.length();
? ? ? ? int lj=S.length();
? ? ? ? int count=0;
? ? ? ? for(int i=0;i<li;i++)
? ? ? ? {
????????????????for(int j=0;j<lj;j++)
? ? ? ? ? ? ? ? {
? ? ? ? ? ? ? ? ? ? ? ? if(S[j]==J[i])
? ? ? ? ? ? ? ? ? ? ? ? ? ? count++;
????????????????}
????????}
? ? ? ? return count;
????}
}
時間復(fù)雜度O(MN)
LeetCode上最優(yōu)解
int numJewelsInStones(stringJ,stringS){
????int res =0;
????set?<char> setJ(J.begin(), J.end());
????for(chars : S)
????????if(setJ.count(s))?
????????????res++;
????return res;?
?}
此方法可以做到O(N)
此方法使用到了set方法，若要使用，需引入#include<set>
屬于c++中的stl部分，不熟悉
