771. Jewels and Stones

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.

时间复杂度 O(mn)

题目已表示 J 中字母不会重复，就不用考虑循环对比重复计数的问题。所以一开始就想到时间复杂度 O(mn)。字符串字符和字符循环遍历比较。

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 (J.charAt(i) == S.charAt(j)) {
                num++;
            }
        }
    }
    return num;
}

上面方法有点傻的地方，比如 J 的第 1 位字母和 S 第 1 位字母相同，按理说 S 第 1 位字母就没有再比较的必要了。但是上面的方法是每个都比较一遍。

时间复杂度 O(m+n)

Explanation

1. read J and build jewels hash set.
2. read S and count jewels.

Time complexity
I used hash set and it's O(1) to check if it contains an element.
So the total time complexity will be O(M+N), instead of O(MN)

public int numJewelsInStones(String J, String S) {
    int res = 0;
    Set setJ = new HashSet();
    for (char j: J.toCharArray()) setJ.add(j);
    for (char s: S.toCharArray()) if (setJ.contains(s)) res++;
    return res;
}