@chawuciren
2018-11-22 14:46
字数 825
阅读 610
leetcode
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.
J里面的元素一定是不同的,所以我就想,从J里面每次取一个元素,然后和S的每一个元素进行比较,相同就加一,然后进行下一个元素的比较,完美,没毛病。
int numJewelsInStones(char* J, char* S){
int count=0;
int len1=strlen(J);
int len2=strlen(S);
for(int i=0;i<len1;i++){
for(int j=0;j<len2;j++){
if(J[i]==S[j]){
count+=1;
}
}
}
return count;
}
Runtime: 4 ms, faster than 100.00% of C online submissions for Jewels and Stones.
真的easy