LeetCode 771. Jewels and Stones

You're given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels.

Letters are case sensitive, so "a" is considered a different type of stone from "A".

吧

Example 1:

Input: jewels = "aA", stones = "aAAbbbb"
Output: 3

Example 2:

Input: jewels = "z", stones = "ZZ"
Output: 0

Constraints:

• 1 <= jewels.length, stones.length <= 50
• jewels and stones consist of only English letters.
• All the characters of jewels are unique.

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就是判断一个字符串里的字符有多少个出现在了另一个字符串里……没啥意思……就把jewels变成set然后遍历stones看在不在set里……

class Solution {
    public int numJewelsInStones(String jewels, String stones) {
        Set<Character> set = new HashSet<>();
        for (char c : jewels.toCharArray()) {
            set.add(c);
        }
        int count = 0;
        for (char c : stones.toCharArray()) {
            if (set.contains(c)) {
                count++;
            }
        }
        return count;
    }
}

也可以直接用indexOf……

class Solution {
    public int numJewelsInStones(String jewels, String stones) {
        int count = 0;
        for (char c : stones.toCharArray()) {
            if (jewels.indexOf(c) != -1) {
                count++;
            }
        }
        return count;
    }
}

还有人用regex的……Loading...