class Solution { //方法一 public int singleNumber(int[] nums) { Set<Integer> set = new HashSet<>(); for (int num: nums){ if(set.contains(num)){ set.remove(num); }else{ set.add(num); } } Iterator<Integer> iterator = set.iterator(); return iterator.next(); } return -1; ...
1: intsingleNumber(int A[], int n) { 2: int left = A[0]; 3:for(int i =1;i< n;i++) 4: { 5: left = left ^ A[i]; 6: } 7: return left; 8: }
AI代码解释 classSolution{publicintsingleNumber(int[]nums){Map<Integer,Integer>map=newHashMap<>();for(int num:nums){Integer count=map.get(num);//get() 方法获取元素不存在时返回nullcount=count==null?1:++count;//count为null 时证明元素不存在,则频率改为1,否则count频率+1map.put(num,count);...
classSolution {public:intsingleNumber(vector<int>&nums) {if(nums.empty())return0;intret = nums[0];for(inti =1; i < nums.size(); i ++) ret^=nums[i];returnret; } }; 解法三:先排序,再遍历找出孤异元素 classSolution {public:intsingleNumber(vector<int>&nums) { sort(nums.begin(), ...
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, 要找出只出现了一次的数我们想到可以用XOR, 代码如下: publicclassSolution {publicintsingleNumber(int[] a) {intresult = a[0];for(inti=1;i<a.length;i++){ result^=a[i];
Given an array of integers, every element appearsthreetimes 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:开辟map记录次数 classSolution {public:intsingleNumber(vector<int>&nums) { ...
classSolution {public:intsingleNumber(vector<int>&nums) { unordered_set<int>st;for(intnum : nums) {if(st.count(num)) st.erase(num);elsest.insert(num); }return*st.begin(); } }; Java 解法一: classSolution {publicintsingleNumber(int[] nums) { ...
Solution classSolution{public:intsingleNumber(vector<int>& nums){intres = nums[0];intsize = nums.size();for(inti =1; i < size; i++) { res ^= nums[i]; }returnres; } }; 解题描述 这道题题意是,给出一个数组,数组中有一个数字只会出现一次,其他所有数字都会出现两次。另外,解法上题目...
publicclassSolution {publicintsingleNumber(int[] nums) {if(nums==null|| nums.length==0){return0; }intsingleNumber=nums[0];for(inti=1;i<nums.length;i++){ singleNumber=singleNumber ^nums[i]; }returnsingleNumber; } } 以下内容摘自他人博客 ...
classSolution{public:vector<int>singleNumber(vector<int>&nums){intn=nums.size();intres=nums[0];for(inti=1;i<n;i++)res^=nums[i];intpos;for(pos=0;pos<31;pos++)if((res&(1<<pos))!=0)break;intnum1=0,num2=0;for(inti=0;i<n;i++){if((nums[i]&(1<<pos))==0)num1^=nu...