题目链接 题目大意:给出一串数组,里面的数都是两个,只有一个数是一个,把这个只有一个的数找出来.时间复杂度最好是线性的,空间复杂度最好为O(1). 法一:利用map,空间换时间,代码如下(耗时26ms): public int singleNumber(int[] nums) { Map<Integer, Integer> map = new HashMap<Integer, Integer>(); int res = -1; for(int i = 0; i < 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…
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…
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…
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.…
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…
