LeetCode-771. Jewels and Stones
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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | 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
1 2 3 4 5 6 7 8 9 10 11 12 13 | 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; } |
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步