5708. 统计一个数组中好对子的数目
给你一个数组 nums
,数组中只包含非负整数。定义 rev(x)
的值为将整数 x
各个数字位反转得到的结果。比方说 rev(123) = 321
, rev(120) = 21
。我们称满足下面条件的下标对 (i, j)
是 好的 :
0 <= i < j < nums.length
nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])
请你返回好下标对的数目。由于结果可能会很大,请将结果对 109 + 7
取余 后返回。
示例 1:
输入:nums = [42,11,1,97] 输出:2 解释:两个坐标对为: - (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 。
示例 2:
输入:nums = [13,10,35,24,76] 输出:4
提示:
1 <= nums.length <= 105
0 <= nums[i] <= 109
nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])=>
(nums[i] - rev(nums[i])) == (nums[j] - rev(nums[j]))=>
Transform each nums[i] into (nums[i] - rev(nums[i])). Then, count the number of (i, j) pairs that have equal values.=>
add c*(c-1)/2 to the ans for each c in the Counter()
python
class Solution: def countNicePairs(self, nums: List[int]) -> int: for i in range(len(nums)): nums[i]=nums[i]-int(str(nums[i])[::-1]) cnt=collections.Counter(nums) ans=0 for c in cnt.values(): ans+=(c*(c-1)/2) return int(ans)%(10**9+7)
C++
class Solution { public: int countNicePairs(vector<int>& nums) { long ans=0; unordered_map<int,int> m; for(int n:nums){ string s=to_string(n); reverse(s.begin(),s.end()); ans+=(m[n-stoi(s)])++; } return ans%(long)(1e9+7); } };