Question Given an array of integers, every element appears twice except for one. Find that single one. Solution 1 -- Set We can use a hash set to record each integer's appearing time. Time complexity O(n), space cost O(n) public class Solution { publ…

136. Single Number -- Easy 解答 相同的数,XOR 等于 0,所以,将所有的数字 XOR 就可以得到只出现一次的数 class Solution { public: int singleNumber(vector<int>& nums) { int s = 0; for(int i = 0; i < nums.size(); i++) { s = s ^ nums[i]; } return s; } }; 参考 LeetCode Problems' So…

异或巧用:Single Number 今天刷leetcode,碰到了到题Single Number.认为解答非常巧妙,故记之... 题目: Given an array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it w…

原题目 Given a non-empty array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 普通解 刚看到的这个题目第一时间想到的就是a[i]++,最后…

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. For example: Given nums = [1, 2, 1, 3, 2, 5], return [3, 5]. Note: The order…

Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 这道题是之前那道 Single Number 单独的数字的延伸,那道题的解法…

Given an array of integers, every element appears twice except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 本来是一道非常简单的题,但是由于加上了时间复杂度必须是O(n),并且空间复杂度为O(1)…

[1]LeetCode 136 Single Number 题意:奇数个数,其中除了一个数只出现一次外,其他数都是成对出现,比如1,2,2,3,3...,求出该单个数. 解法:容易想到异或的性质,两个相同的数异或为0,那么把这串数从头到尾异或起来,最后的数就是要求的那个数. 代码如下: class Solution { public: int singleNumber(vector<int>& nums) { ; ;i<nums.size();i++) sum ^= nums[i…

问题: Given an array of integers, every element appears three times except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?   Single Number I 升级版,一个数组中其它数出现了…

Given 2*n + 1 numbers, every numbers occurs twice except one, find it. Have you met this question in a real interview? Yes Example Given [1,2,2,1,3,4,3], return 4 Challenge One-pass, constant extra space. LeetCode上的原题,请参见我之前的博客Single Number. class So…