Leetccode 136 SingleNumber 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? 1.自己想到了先排序,再遍历数…
136. Single Number 意思就是给你一堆数,每个数都出现了两次,只有一个数只出现了一次,找出这个数 位运算(和c艹一样) &:按位与 |:按位或 ^:异或(一样为0,不一样为1) 再说一下异或的性质,满足交换律和结合律 因此: 对于任意一个数n n ^ 0 = n n ^ n = 0 对于这道题来说,所有数依次异或剩下的就是那个数了 class Solution(object): def singleNumber(self, nums): ans = 0 for i in nums…
https://leetcode.com/problems/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?…
The question: 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? My analysis: Thi…
题目描述(面试常考题) 借助了异或的思想 class Solution { public: int singleNumber(vector<int>& nums) { ; ; i < nums.size() ; i++) ret = ret ^ nums[i]; return ret; /* 出现3次的情况借助了位运算的思想,相同的数 则 在int型的各32位都相等,所以32位中各位求和取余就可以得到结果 */ class Solution { public: int singl…
1.题目意思:在数组中,只有一个数字只出现了一次 其他的都出现了两次.找出那个只出现一次的数字. //利用位运算 异或 两个相同的数字异或为0 public int singleNumber(int[] nums) { int res = 0 ; for (int i = 0; i < nums.length; i++) { res ^=nums[i]; } return res; } 2.题目意思:数组中,每个数字都出现三次 除了只有一个数字只出现了一次.找出那个只出现一次的数字. // 方…
First. 陈列一下“异或”的一些性质 异或是一种基于二进制的位运算,用符号XOR或者 ^ 表示,其运算法则是对运算符两侧数的每一个二进制位,同值取0,异值取1. 它与布尔运算的区别在于,当运算符两侧均为1时,布尔运算的结果为1,异或运算的结果为0. 性质 1.交换律 2.结合律(即(a^b)^c == a^(b^c)) 3.对于任何数x,都有x^x=0,x^0=x 4.自反性 A ^ B ^ B = A ^ 0 = A 应用 一.交换两个整数的值而不必用第三个参数 a = 9; b =…
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…
