Leetcode_1512. Number of Good Pa
2020-07-16 本文已影响0人
钱晓缺
关键找到每个元素的出现的次数:
方法一:
res=0
for i in set(nums):
res=nums.count(i)*(nums.count(i)-1)/2+res
return res
方法二:
returnsum(v * (v -1) //2 for v in Counter(nums).values())
关键找到每个元素的出现的次数:
方法一:
res=0
for i in set(nums):
res=nums.count(i)*(nums.count(i)-1)/2+res
return res
方法二:
returnsum(v * (v -1) //2 for v in Counter(nums).values())