771. Jewels and Stones

最新推荐文章于 2020-12-27 23:49:46 发布

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最新推荐文章于 2020-12-27 23:49:46 发布

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分类专栏： LeetCode 字符串处理

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本文介绍了一种算法问题，即给定两个字符串J和S，分别代表宝石类型和拥有的石头，任务是找出S中有多少个石头也是J中的宝石。文章提供了两种解决方案：一种是通过两次for循环遍历字符串实现；另一种则是利用哈希表提高查找效率。

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You're given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in Sis 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.

方法一：两次for循环，对字符串遍历复杂度 O(i*j)

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

方法二：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

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