LWC 69: 771. Jewels and Stones

LWC 69: 771. Jewels and Stones

传送门：771. Jewels and Stones

Problem:

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.

思路：

记录J的信息，遍历S，采用Hashmap的思路，每次查询O(1)，总时间复杂度为O(n)

Java版本：

public int numJewelsInStones(String J, String S) {
int[] count = new int[64];
for (char c : J.toCharArray()) {
count[c - 'A']++;
}
int ans = 0;
for (char c : S.toCharArray()) {
if (count[c - 'A'] >= 1) ans ++;
}
return ans;
}

Python版本：

class Solution(object):
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
info = set(J)
ans = 0
for c in S:
if c in info: ans += 1
return ans

再精简一波：

class Solution(object):
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
info = set(J)
return sum(c in info for c in S)