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- 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;
}
};