1.数组中仅仅有一个元素仅仅出现一次,其余出现偶数次. 利用异或中同样元素相互抵消的方式求解. 2.数组中仅仅有2个元素仅仅出现一次.其余出现偶数次. class Solution { public: int singleNumber(int A[], int n) {//返回仅仅出现一次的元素 if(n==1)return A[0]; int i,sum=A[0]; for(i=1;i<n;++i)sum^=A[i]; return sum; } vector<int> twoSing…

题目要求 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? 一个int数组中,除了有一个int只出现了一次,其他int都出现了两次,请找出…

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 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…

题目地址:here 题目大意:一个整数数组中,只有一个数出现一次,其余数都出现3次,在O(n)时间,O(1)空间内找到这个出现一次的数 对于”只有一个数出现一次,其余数出现2次“的情况,很简单,只要把数组中所有数异或的结果就是这个出现一次的数,leetcode上也有这个题目. 关于本题:我们用一个大小为32的数组cnt来记录:(int数据的每个二进制位1的出现次数)%3,程序中cnt[i]记录第i+1个二进制位.如果某个数出现了三次并且他的第i个二进制位为1,因为做了模3操作,那么该数对数组cn…

原题链接在这里:https://leetcode.com/problems/single-number-iii/ 题目: 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: Giv…

原题地址:http://oj.leetcode.com/problems/single-number-ii/ 题意:Given an array of integers, every element appears three times except for one. Find that single one. 要求:和single number一样,线性时间复杂度,不能使用额外空间. 解题思路:这道题就比较难了.也是考察位操作.网上的位操作解法看了好半天也没有得其精髓.由于序列中除了那唯一的…

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…

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…

I title: 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:异或 class Solution { public: int…

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…

Single Number Given an array of integers, every element appears twice except for one. Find that single one. def singleNumber(self, A): l = len(A) if l < 2: return A[0] A.sort() for i in range(0,l-1,2): if A[i] != A[i+1]: return A[i] return A[l-1] 思路:…

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? 题意:给定数组,除一个数仅出现一次外,其余都出现三次,找到仅出现一次的数字…

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? 题意:给定一未排序的数组,其中只有一个数仅出现一次,其余都出现两次,找到仅出现一次的这…

还记得<剑指offer>和<编程之美>等书上多次出现的找一个数组中仅仅出现一次的数那个题吗? leetcode也有这道题 链接here  相信大家都知道用异或在O(n)的时间复杂度内求出的方法,这里不再赘述. 以下就是上题的升级版 Given an array of integers, every element appears three times except for one. Find that single one. Note: Your algorithm should…

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? 感谢cnblog博主feiling,这篇博客一方面参考了feiling的博客,也加入了…

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? Hide Tags Bit Manipulation   数组中的数均出现3…

问题描述: 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] Ou…

Given an array of integers, every element appears twice except for one. Find that single one. 非常简单的一道题. 直接相异或剩下的那个数就是答案.原理是两个相等的数异或的值为0. class Solution { public: int singleNumber(int A[], int n) { int temp; ;i!=n;i++) temp=temp^A[i]; return temp; } }…

问题: 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?   分析: 数组中的数除了一个只出现了一次之外,其它都出现了两次, 要找出只出…

题目: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? 思路:一个比较好的方法是从位运算来考虑.每一位是0或1,由于除了一个数…

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? #-------------------------------------------…

这个系列一共有三题,第一题是一组数里除了一个数出现一次之外,其他数都是成对出现,求这个数. 第二题是一组数里除了两个数出现一次外,其他数都是成对出现,求这两个数 第三题是一组数里除了一个数出现一次外,其他数都是出现三次,求这个数. 先说第一题,这题很简单,就是将所有的数全部异或一遍,由于两个相同的数异或之后得到0,所以最后得到的结果就是要求的那个数. 第二题有点难, 解题方法是,设所求的那两个数为n1,n2.那么n1和n2必然不相等,那么它们必然至少有一个bit的值不等. 那么与题1一样,将所有…

题意:给一个序列,序列中只有1个是单个的,其他都是成对出现的.也就是序列中有奇数个元素.要求找出这个元素. 思路:成对出现用异或最好了.两个同样的数一异或就变零,剩下的,就是那个落单的. class Solution { public: int singleNumber(vector<int>& nums) { ; ; i<nums.size(); i++) res^=nums[i]; return res; } }; AC代码…

#question : 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? class Solution(object): def si…

Description: Given an array of integers, every element appears three times except for one. Find that single one. 只有一个出现一次的数字,其他的都出现了3次,找出出现一次的那个数字. public class Solution { public int singleNumber(int[] nums) { Map<Integer, Integer> map = new HashMap…

Description: Given an array of integers, every element appears twice except for one. Find that single one. 找出只出现一次的数. public class Solution { public int singleNumber(int[] nums) { Map<Integer, Integer> map = new HashMap<Integer, Integer>(); fo…

题目描述: 给定一个包含n个整数的数组,除了一个数出现一次外所有的整数均出现三次,找出这个只出现一次的整数. 题目来源: http://oj.leetcode.com/problems/single-number-ii/ 题目分析: 对于除出现一次之外的所有的整数,其二进制表示中每一位1出现的次数是3的整数倍,将所有这些1清零,剩下的就是最终的数. 用ones记录到当前计算的变量为止,二进制1出现“1次”(mod 3 之后的 1)的数位.用twos记录到当前计算的变量为止,二进制1出现“2次”(…

题意: 给一个数组,其中仅有两个元素是出现1次的,且其他元素均出现2次.求这两个特殊的元素? 思路: 跟查找单个特殊的那道题是差不多的,只是这次出现了两个特殊的.将数组扫一遍求全部元素的异或和 x,结果也就是这两个特殊的元素的异或和了.现在必须找到两个当中的一个,才能恢复出另外一个.注意到x的二进制位中必定有1,那么只需要找到其中任意一个1对应的整数,再将数组中所有该位为1的计算他们的异或和a,a就是结果了,而另一个数b=x^a. vector<int> singleNumber(vector…