Single Number I 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? 直接异或解决. class Solution { pu…
Single Number: 1. 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? 代码: class Solution { publi…
Single Number I : 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? Solution: 解法不少,贴一种: class…
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:…
[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…
Given an array of integers, every element appears twice except for one. Find that single one. 本题利用XOR的特性, X^0 = X, X^X = 0, 并且XOR满足交换律. class Solution(object): def singleNumber(self, nums): """ :type nums: List[int] :rtype: int ""…
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…
Problem statement Elementary knowledge: There is a popular question when I seeked my job at Beijing: swap the value of two numbers without allocate a new memory. Normally, there are two solutions: 1 int main(){ 2 int a = 3; 3 int b = 5; 4 5 a += b; 6…
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…
