[抄题]: 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…
[抄题]: 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? [暴力解法]: 时间分析:n 空间分析:新建n的数组来存储 [奇葩输出条件]…
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…
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…
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 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…
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.…