LeetCode - Number of Good Pairs Given an array of integers nums. A pair (i,j) is called goodifnums[i] == nums[j] and i <j. Return the number of good pairs. Example1: Input: nums= [1,2,3,1,1,3] Output:4Explanation: There are4 good pairs (0,3), (0,4), (3,4), (...
题目 1512. Number of Good Pairs 解题方法 先构造计数字典,返回值rat,然后遍历字典根据其中每个数字的出现次数套用公式n(n-1)/2即可。 时间复杂度:O(n) 空间复杂度:O(n) 代码 classSolution:defnumIdenticalPairs(self,nums:List[int])->int:dic,rat=collections.Counter(nums),0forkey in dic.keys():ra...
1512. Number of Good Pairs # 题目 # Given an array of integers nums. A pair (i,j) is called good if nums[i] == nums[j] and i < j. Return the number of good pairs. Example 1: Input: nums = [1,2,3,1,1,3] Output: 4 Explanation: There are 4 good pairs (0,3
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good_pairs = 0 # For each number in nums1, find all its divisors and check against the hashmap for num in nums1: for j in range(1, int(num ** 0.5) + 1): if num % j == 0: # j is a divisor if j in divisor_count: ...
Leetcode solution 1512. Number of Good Pairs Blogger: https://blog.baozitraining.org/2020/07/leetcode-solution-1512-number-of-good.html Youtube: https://youtu.be/dvnjvOLh88k 博客园:https://www.cnblogs.com/baozitraining/p/13338525.html ...
[leetcode] 1512. Number of Good Pairs Description Given an array of integers nums. A pair (i,j) is called good if nums[i] == nums[j] and i < j. Return the number of good pairs. Example 1: Input: nums = [1,2,3,1,1,3]...
Runtime:1 ms, faster than81.21%of Java online submissions for Number of Good Pairs. Memory Usage:39.7 MB, less than14.56%of Java online submissions for Number of Good Pairs. 整体表现不太好。只能说一般。
【leetcode】1295. Find Numbers with Even Number of Digits 2019-12-23 10:45 − 题目如下: Given an array nums of integers, return how many of them contain an even number of digits. Example 1: Input: nums = [12,345... seyjs 0 668 leetcode 24. Swap Nodes in Pairs 2019-12...
pairs,数对; difference,差; absolute difference,差值的绝对值; absolute difference K,差值的绝对值=K。 题目:Given an integer arraynumsand an integerk, returnthe number of pairs(i, j)wherei < jsuch that|nums[i] - nums[j]| == k.