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
andJ
will consist of letters and have length at most 50.- The characters in
J
are distinct.
public int numJewelsInStones(String J, String S) {//my Map<Character,Integer> map = new HashMap<Character, Integer>(); int re=0; for (int i = 0; i < J.length(); i++) { map.put(J.charAt(i),0); } for (int i = 0; i < S.length(); i++) { if(map.containsKey(S.charAt(i))){ map.put(S.charAt(i),map.get(S.charAt(i))+1); } } for (Map.Entry entry:map.entrySet()) { re += (Integer) entry.getValue(); } return re; }
简洁版,使用Set
public int numJewelsInStones(String J, String S) {//my int re=0; Set<Character> set = new HashSet<Character>(); for (int i = 0; i < J.length(); i++) { set.add(J.charAt(i)); } for (int i = 0; i < S.length(); i++) { if(set.contains(S.charAt(i))){ re++; } } return re; }