LeetCode Problem -- 771. Jewels and Stones

版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/m0_38088298/article/details/85344966
  • 描述: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

  • 分析:给定一个字符串代表宝石,另一个字符串代表石头,要求检测石头中有多少宝石,其实就是检测J字符串字母在S字符串中的出现的数目,需要注意的是检测的是单个字母而不是J整个字符串。
  • 思路一:用map对石头的种类和数目进行统计,之后用J中的字母去查找map就可以了。
class Solution {
public:
    int numJewelsInStones(string J, string S) {
        map<char, int> stone_box;
        int jewels_num = 0;
        for (int i = 0; i < S.size(); i++) {
            if (stone_box.find(S[i]) == stone_box.end()) {
                stone_box[S[i]] = 1;
            } else {
                stone_box[S[i]] += 1;
            }
        }
        for (int i = 0; i < J.size(); i++) {
            if (stone_box.find(J[i]) != stone_box.end()) {
                jewels_num += stone_box.find(J[i]) -> second;
            }
        }
        return jewels_num;
    }
};

猜你喜欢

转载自blog.csdn.net/m0_38088298/article/details/85344966