差130个样例，等佬解

class Solution:
    def ifqual(self,str1,str2):
        return int(str1)==int(str2)
    def change(self,str1,str2):
        str1 = list(str1)
        n=len(str1)
        t=0
        for i in range(n):
            for j in range(i+1,n):
                str1[i],str1[j]=str1[j],str1[i]
                t+=1
                if self.ifqual("".join(str1),str2):
                    return True
                if t==2:
                    str1[i],str1[j]=str1[j],str1[i]   
        return False
         
    def countPairs(self, nums: List[int]) -> int:
        n=len(nums)
        ret=0
        for i in range(n):
            for j in range(i+1,n):
                if nums[i]==nums[j]:
                    ret+=1
                else:
                   num1=str(nums[i])
                   num2=str(nums[j])
                   if self.change(num1,num2) or self.change(num2,num1):
                       ret+=1
        return ret

大佬题解

class Solution:
    def countPairs(self, nums: List[int]) -> int:
        nums.sort()
        ret=0
        dict=defaultdict(int)
        for num in nums:
            #不交换
            st={num}
            #num的长度
            s=list(str(num))
            m=len(s)
            #数位交换
            for i in range(m):
                for j in range(i+1,m):
                    s[i],s[j]=s[j],s[i]
                    st.add(int(''.join(s)))
                    for p in range(i+1,m):
                      for q in range(p+1,m):
                        s[p],s[q]=s[q],s[p]
                        st.add(int(''.join(s)))
                        #交换2次
                        s[p],s[q]=s[q],s[p]
                    #换回来
                    s[i],s[j]=s[j],s[i]
            ret+=sum(dict[v] for v in st)
            dict[num]+=1
        return ret         

思路确实好用，2题都能解。

ps:

力扣刷完还能atcoder>codeforces.