Topic Source: https://leetcode.com/problems/jewels-and-stones/
Problem Description: two strings J and S, S appeared to find out how many times a character string which contains J's.
for example:
String J | String S | result |
---|---|---|
aA | aAAbsdfe | 3 |
b | Brbaaaa | 0 |
solution
- Double traverse two strings, comprising checking whether the character substring, the time complexity of Ο (n ^ 2)
public int numJewelsInStones(String J, String S) {
int sum = 0;
char[] jChar = J.toCharArray();
char[] sChar = S.toCharArray();
int jLength = jChar.length;
int sLength = sChar.length;
for(int i=0;i<jLength;i++){
for(int j=0;j<sLength;j++){
if(jChar[i] == sChar[j]) {
sum ++;
}
}
}
return sum;
}
}