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 and J 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;
    }

  

原文地址:https://www.cnblogs.com/zhacai/p/10429670.html