1814. Statistics the number of good pairs in an array One question daily

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1814. Statistics the number of good pairs in an array
medium
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Superior Uber
Give you an array nums ，The array contains only non -negative integers。definition rev(x) The value is an integer x The results obtained by the digital bits are reversed。for example rev(123) = 321 ， rev(120) = 21 。We call the lattering pair of the following conditions (i, j) yes OK ：

0 <= i < j < nums.length
nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])
Please return the number of bidding pairs。Because the result may be great，Please right 109 + 7 Take a surplus Back。

 

Exemplary example 1：

enter：nums = [42,11,1,97]
Output：2
explain：Two coordinates right：
 - (0,3)：42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121 。
 - (1,2)：11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12 。
Exemplary example 2：

enter：nums = [13,10,35,24,76]
Output：4
 

hint：

1 <= nums.length <= 105
0 <= nums[i] <= 109

Thinking

这道题没想到yesHash table，还以为yesDouble pointer。
It has been done for a long time，Unexpectedly，Just look at the problem and solve the problem，I also understand。
总之就yes，好对子其实就yes two i - res(i) The same number!
Word gamesa feeling of：

42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121 。
不就yes 42-res(42) == 97 - res(97) ?？

when*cnt[i-res(i)]*When there is no existence，Will return0.

ans += cnt[i-j]

Thus, incnt[i-j]have2次值的hour候才Will return到ansmiddle
becausecnt[i-j]=1hour，yes不会经过ansAssociated

cnt[i-j] += 1

solution：

·ElementnumsW - rev(num[il)
·Calculate the number of times per element
·Calculate the difference between the number of times each time
@Lazy

---

code show as below：

class Solution:
    def countNicePairs(self, nums: List[int]) -> int:
        def res(num: int) -> int:
            return int(str(num)[::-1])
        cnt = Counter()
        ans = 0
        for i in nums:
            j = res(i)
            # whencnt[i-res(i)]When there is no existence，Will return0.
            ans += cnt[i-j]
            # Thus, incnt[i-j] have2次值的hour候才Will return到ansmiddle
            # becausecnt[]=1hour，yes不会经过ansAssociated
            cnt[i-j] += 1
        return ans % (10 ** 9 + 7)