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.
方案一：O（n^2)
class Solution {
public int numJewelsInStones(String J, String S) {
int num = 0;
for(int i = 0; i < J.length(); i++) {
for(int j = 0; j < S.length(); j++) {
if(S.charAt(j) == J.charAt(i)) num++;
}
}
return num;
}
}
Do you have a more effective solution ?
方案二：时间复杂度O（n）
class Solution {
public int numJewelsInStones(String J, String S) {
int num = 0;
Map<String, Integer> map = new HashMap<String, Integer>();
//将S转成 map<char, int>形式
for(int i = 0; i < S.length(); i++) {
if(map.get(String.valueOf(S.charAt(i))) == null) {
map.put(String.valueOf(S.charAt(i)), 1);
}else {
if(map.get(String.valueOf(S.charAt(i))) != 0 ) {
map.put(String.valueOf(S.charAt(i)), map.get(String.valueOf(S.charAt(i))) + 1);
};
}
}
for(int j = 0; j < J.length(); j++) {
num = num + map.get(String.valueOf(J.charAt(j)));
}
return num;
}
}