第二次计数,同样声明一个长度为2001的int数组count2,如果count中的当前元素不等于0,就将当前元素作为count2数组的索引,元素值累加,加完后判断元素值是否大于等于2,如果大于,直接返回false。 publicbooleanuniqueOccurrences2(int[] arr){int[] count =newint[2001];for(intnum : arr) { count[num+1000]++; }...
LeetCode.1207-唯一的元素出现次数(Unique Number of Occurrences) 这是小川的第419次更新,第452篇原创 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第269题(顺位题号是1207)。给定一个整数数组arr,当且仅当该数组中每个元素的出现次数唯一时,返回true。 例如: 输入:arr = [1,2,2,1,1,3] 输出:...
https://leetcode-cn.com/problems/unique-number-of-occurrences/耗时解题:22 min 题解:4 min题意给你一个整数数组 arr,请你帮忙统计数组中每个数的出现次数。如果每个数的出现次数都是独一无二的,就返回 true;否则返回 false。思路两个哈希表,一个用来离散统计出现次数,第二个测试出现次数是否独一无二。
1typedeflonglongll;2#define_for(i,a,b) for(int i = (a);i < b;i ++)3#define_rep(i,a,b) for(int i = (a);i > b;i --)4#defineINF 0x3f3f3f3f5#definepb push_back6#definemaxn 1000078classSolution9{10public:11booluniqueOccurrences(vector<int>&arr)12{13intHash[2003];14mem...
Given an array of integers arr, write a function that returns true if and only if the number of occurrences of each value in the array is unique. Example 1: Input: arr = [1,2,2,1,1,3] Output: true Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two va...
Can you solve this real interview question? Unique Number of Occurrences - Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise. Example 1: Input: arr = [1,2,2,1,1,3] Outpu
1207. Unique Number of Occurrences* 1207. Unique Number of Occurrences* https://leetcode.com/problems/unique-number-of-occurrences/ 题目描述 Given an array of integers arr, write a function that returns true if and only if the number of ...
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今天介绍的是LeetCode算法题中Easy级别的第269题(顺位题号是1207)。给定一个整数数组arr,当且仅当该数组中每个元素的出现次数唯一时,返回true。 例如: 输入:arr = [1,2,2,1,1,3] 输出:true 说明:值1出现3次,值2出现2次,值3出现1次。没有两个值出现的次数相同。
Given an array of integersarr, write a function that returnstrueif and only if the number of occurrences of each value in the array is unique. Example 1: Input: arr = [1,2,2,1,1,3] Output: true Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values...