1classSolution {2publicString findDifferentBinaryString(String[] nums) {3intlen = nums[0].length();4HashSet<String> set =newHashSet<>();5for(String num : nums) {6set.add(num);7}89intcount = (int) Math.pow(2, len);10for(inti = 0; i < count; i++) {11StringBuilder sb =ne...
Leetcode 1304.和为零的N个唯一整数(Find N Unique Integers Sum up to Zero) Leetcode 1304.和为零的N个唯一整数 1 题目描述(Leetcode题目链接) 给你一个整数 n,请你返回 任意 一个由 n 个 各不相同 的整数组成的数组,并且这 n 个数相加和为 0 。 提示:1 <= n <= 1000 2 题解 ...
LeetCode 1304. Find N Unique Integers Sum up to Zero 原题链接简单 作者: wzc1995 , 2020-01-01 09:52:23 , 所有人可见 , 阅读 881 3 题目描述 给你一个整数 n,请你返回 任意 一个由 n 个各不相同 的整数组成的数组,并且这 n 个数相加和为 0。 样例 输入:n = 5 输出:[-7,-1,1...
Can you solve this real interview question? Find N Unique Integers Sum up to Zero - Given an integer n, return any array containing n unique integers such that they add up to 0. Example 1: Input: n = 5 Output: [-7,-1,1,3,4] Explanation: These array
publicint[]sumZero(intn){int[] A =newint[n];for(inti =0; i < n; ++i) A[i] = i *2- n +1;returnA; } 这个solution是通过找对称性规律直接在Array[i]里赋相反值达到的效果 评论里提出我的代码n==2时返回的数组是{0,0},这违反了题目中的一致性。
1304. Find N Unique Integers Sum up to Zero # 题目 # Given an integer n, return any array containing n unique integers such that they add up to 0. Example 1: Input: n = 5 Output: [-7,-1,1,3,4] Explanation: These arrays also are accepted [-5,-1,1,2,3] ,
这次有点快,有点不可思议,主要是看这个数据是奇数项还是偶数项; Runtime:0 ms, faster than100.00%of Java online submissions for Find N Unique Integers Sum up to Zero. Memory Usage:42.8 MB, less than6.64%of Java online submissions for Find N Unique Integers Sum up to Zero. ...
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[leetcode] 1304. Find N Unique Integers Sum up to Zero,DescriptionGivenanintegern,returnanyarraycontainingnuniqueintegerssuchthattheyaddupto0.Example1:Input:n=5Output:[-7,-1,1,3,4]Explanation:Thesea
The number of files given is in the range of [1,20000]. You may assume no files or directories share the same name in the same directory. You may assume each given directory info represents a unique directory. Directory path and file info are separated by a single blank space. ...