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())
关键找到每个元素的出现的次数： 方法一： res=0 for i in set(nums): ...
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本文标题：Leetcode_1512. Number of Good Pa
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