771.Jewels and Stones
1、题目描述
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
Note:
S and J will consist of letters and have length at most 50.
The characters in J are distinct.
解法(1)
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
if len(J)>50 or len(S)>50:
return 0
num=0
for i in range(len(J)):
for j in range(len(S)):
if J[i]==S[j]:
num+=1
return num
解法(2)
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
return sum(map(J.count,S))
map(A,B) 会根据提供的函数对指定序列做映射。
第一个参数 function 以参数序列中的每一个元素调用 function 函数,返回包含每次 function 函数返回值的新列表。
map(square, [1,2,3,4,5]) # 计算列表各个元素的平方
[1,4,9,16,25]
()