771.Jewels and Stones (leetcode)

Jewels and Stones

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题目

leetcode题目

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.

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解决

1.通过遍历字符串S和J，两两进行比较，判断stone中有多少颗jewel。

• 时间复杂度为O(s * j)。(s为字符串S的长度，j为字符串J的长度)
• 空间复杂度为O(1)。

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        int result = 0;
        int jlen = J.length();
        int slen = S.length();
        for (int i = 0; i < slen; i++) {
            for (int j = 0; j < jlen; j++) {
                if (S[i] == J[j]) {
                    result++;
                    break;
                }
            }
        }
        return result;
    }
};

2.通过hash的方法，先将字符串J中的每个字符串当作key存起来，再遍历字符串S，检查是否存在这些key。

• 时间复杂度为O(s + j)。(s为字符串S的长度，j为字符串J的长度)
• 空间复杂度为O(j)。

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        int result = 0;
        int jlen = J.length();
        int slen = S.length();
        unordered_map<char, int> stone;
        for (int i = 0; i < jlen; i++) {
            stone[J[i]] = 1;
        }
        for (int i = 0; i < slen; i++) {
            // 检查是否存在对应的key
            if (stone.find(S[i]) != stone.end()) {
                result++;
            }
        }
        return result;
    }
};