Linked Url:https://leetcode.com/problems/single-number/ 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 withou…

leetcode 136. Single Number 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? 解题思路: 如果以线性复杂度和…

LeetCode 136. Single Number(只出现一次的数字)…

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…

136. Single Number 除了一个数字,其他数字都出现了两遍. 用亦或解决,亦或的特点:1.相同的数结果为0,不同的数结果为1 2.与自己亦或为0,与0亦或为原来的数 class Solution { public: int singleNumber(vector<int>& nums) { if(nums.empty()) ; ; ;i < nums.size();i++) res ^= nums[i]; return res; } }; 137. Single N…

翻译 给定一个整型数组,除了某个元素外其余元素均出现两次. 找出这个仅仅出现一次的元素. 备注: 你的算法应该是一个线性时间复杂度. 你能够不用额外空间来实现它吗? 原文 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 yo…

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? Example 1: Input: [2,2,1] Output:…

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? 题目标签:Hash Table 题目给了我们一个 nums array, array 里…

原题地址https://leetcode.com/problems/single-number/ 题目描述Given an array of integers, every element appears twice except for one. Find that single one. 给出一个整数数组,除了某个元素外所有元素都出现两次.找出仅出现一次的数字. Note: 注意: Your algorithm should have a linear runtime complexity.…

1.题目大意 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? 给定一个数组的整数,数组中的每个元素都出现了两次.例外地,有一个元素只出现…

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? 求出数组中只出现了一次的数(其他的数都出现了2次) 异或求解. public class…

题目描述: Given an array of integers, every element appears twice except for one. Find that single one. 解题思路: 参考别人的想法,用异或的方法实在是太绝了.A ^ A = 0 and A ^ B ^ A = B,因为只有一个是单着的,那么其他成对的两数相异或的结果是0,最后与单着的数异或还是为这个数. 代码如下: public class Solution { public int singleNu…

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? 思路:将数组元素逐个异或. 举个栗子:2  3  4  2  3 所有数字依次异或运算…

题目要求 Given a non-empty array of integers, every element appears twice except for one. Find that single one. 题目分析及思路 给定一个非空整数数组,除了一个元素只出现一次外,其余元素均出现两次.可以先对数组排序,然后遍历数组,若数组只有一个元素或连续两个元素不相同时则得到返回值,否则则去除这两个元素. python代码 class Solution: def singleNumber(sel…

Given an array of integers, every element appears twice except for one. Find that single one. class Solution { public: int singleNumber(vector<int>& nums) { int size=nums.size(); ||nums.empty()) ; ; ;i<size;++i) res^=nums[i]; return res; } };…

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? 解题思路: 按位异或运算,出现两次的都归零了,剩下的就是出现一次的了,JAVA实现如下…

传送门 Description 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? 思路 题意:给定一个数组,其中只有一个数没有出现两次,要…

最原始的方法:先排序,然后从头查找.若nums[i] = nums[i] + 1则为一对相同的数,i = i  + 2,继续判断.若nums[i] != nums[i] + 1,则输出nums[i].需注意不要越界! class Solution { public: int singleNumber(vector<int>& nums) { int t; ) { t = nums[]; } else { sort(nums.begin(), nums.end()); ; i <…

题意就是从一堆数中找出唯一的一个只存在一个的数.除了那个数,其他的数都是两个相同的数. 通过亦或的性质: 1)a^a = 0 0^a = a 2)交换律 a^b = b^ a 3)结合律 (a^b)^c=a^(b^c) 这样很容易证明将所有的数亦或就能得到唯一的一个只存在一个的数. class Solution { public: int singleNumber(vector<int>& nums) { ]; ;i<nums.size(); ++i){ ans^=nums[i]…

原题地址https://leetcode.com/problems/single-number-ii/ 题目描述Given an array of integers, every element appears three times except for one. Find that single one. 给出一个整数数组,除了某个元素外所有元素都出现三次.找出仅出现一次的数字. Note: 注意: Your algorithm should have a linear runtime co…

翻译 给定一个整型数组,除了某个元素外其余的均出现了三次. 找出这个元素. 备注: 你的算法应该是线性时间复杂度. 你能够不用额外的空间来实现它吗? 原文 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…

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?…

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. Example: Input: [1,2,1,3,2,5] Output: [3,5] Note: The order of the result is…

Question 136. Single Number Solution 思路:构造一个map,遍历数组记录每个数出现的次数,再遍历map,取出出现次数为1的num public int singleNumber(int[] nums) { Map<Integer, Integer> countMap = new HashMap<>(); for (int i=0; i<nums.length; i++) { Integer count = countMap.get(nums…

LeetCode 260. Single Number III(只出现一次的数字 III)…

LeetCode 137. Single Number II(只出现一次的数字 II)…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 异或 字典 日期 [LeetCode] 题目地址:https://leetcode.com/problems/single-number/ Total Accepted: 183838 Total Submissions: 348610 Difficulty: Easy 题目描述 Given a non-empty array of integers…

Single Number 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? 解法一:用map记录每个元素的次数,返回次数为1的元素 cl…

描述: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? 题目要求O(n)时间复杂度,O(1)空间复杂度. 思路1:初步使用暴力搜索,遍历…

Given an array of integers, every element appears twice except for one. Find that single one. 思路: 最经典的方法,利用两个相同的数异或结果为0的性质,则将整个数组进行异或,相同的数俩俩异或,最后得到的就是那个single number,复杂度是O(n) 代码: class Solution { public: int singleNumber(vector<int>& nums) { int…