标签：etc ++ def hat 字母 char s red sel tin
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.J are distinct.给两个字符串，珠宝字符串J和石头字符串S，其中J中的每个字符都是珠宝，S中的每个字符都是石头，区分字母的大小写，问我们S中有多少个珠宝。
Java:
public int numJewelsInStones(String J, String S) {
return S.replaceAll("[^" + J + "]", "").length();
}
Java:
public int numJewelsInStones(String J, String S) {
int res = 0;
Set setJ = new HashSet();
for (char j: J.toCharArray()) setJ.add(j);
for (char s: S.toCharArray()) if (setJ.contains(s)) res++;
return res;
}
Python:
def numJewelsInStones(self, J, S):
return sum(map(J.count, S))
Python:
def numJewelsInStones(self, J, S):
return sum(map(S.count, J))
Python:
def numJewelsInStones(self, J, S):
return sum(s in J for s in S)
Python:
def numJewelsInStones(self, J, S):
setJ = set(J)
return sum(s in setJ for s in S)
Python:
# Time: O(m + n)
# Space: O(n)
class Solution(object):
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
lookup = set(J)
return sum(s in lookup for s in S)
Python: wo
class Solution(object):
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
res = 0
for i in S:
if i in J:
res += 1
return res
C++:
int numJewelsInStones(string J, string S) {
int res = 0;
unordered_set<char> setJ(J.begin(), J.end());
for (char s : S) if (setJ.count(s)) res++;
return res;
}
[LeetCode] 771. Jewels and Stones 珠宝和石头
标签：etc ++ def hat 字母 char s red sel tin
原文地址：https://www.cnblogs.com/lightwindy/p/9834862.html
