771.Jewels and Stones (leetcode)

Jewels and Stones

Jewels and Stones
题目

解决

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.

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;
}
};