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。
Given an array of integersnums, returnthe number ofgood pairs. A pair(i, j)is calledgoodifnums[i] == nums[j]andi<j. Example 1: Input:nums = [1,2,3,1,1,3]Output:4Explanation:There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed. ...
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), (...
1publicintnumIdenticalPairs(int[]nums){2if(nums==null||nums.length==0){3return0;4}56// key: int value; value: number of occurrence7Map<Integer,Integer>lookup=newHashMap<>();8int goodPairsCount=0;9for(int i:nums){10if(lookup.containsKey(i)){11goodPairsCount+=lookup.get(i);12look...
Java实现 classSolution{ publicintnumIdenticalPairs(int[] nums){ intres=0; HashMap<Integer, Integer> map =newHashMap<>(); for(intnum : nums) { intcount=map.getOrDefault(num,0); res += count; map.put(num, count +1); } returnres; ...
Can you solve this real interview question? Number of Good Leaf Nodes Pairs - You are given the root of a binary tree and an integer distance. A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path betwe
Return the number of good leaf node pairs in the tree. Example 1: AI检测代码解析 Input: root = [1,2,3,null,4], distance = 3 Output: 1 Explanation: The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair. ...
class Solution { public int numOfPairs(String[] nums, String target) { int cnt=0; int n=nums.length; for (int i = 0; i < n-1; i++) { for (int j = i+1; j < n; j++) { String temp1=nums[i]+nums[j]; String temp2=nums[j]+nums[i]; if(target.equals(temp1)){ cnt...
新手村100题汇总:王几行xing:【Python-转码刷题】LeetCode 力扣新手村100题,及刷题顺序1 审题题目说明:难度=easy;count number,计数;pairs,数对;difference,差;absolute difference,差值的绝对值;abs…
class Solution: def maxOperations(self, nums: List[int], k: int) -> int: # num_to_cnt[num] 表示 num 还可以使用的次数 num_to_cnt: Dict[int, int] = defaultdict(int) # ans 表示操作次数 ans: int = 0 # 遍历 nums 中的所有数字 for num in nums: if num_to_cnt[k - num] > 0...