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 classSolution{public:intsingleNumber(vector<int>& nums){intres = nums[0];int...
返回的结果也是一个数组,key 为数组中的元素,value 为该元素出现的次数。 functionsingleNumber($nums){$valueCnt=array_count_values($nums);returnarray_search(1,$valueCnt) !==false?array_search(1,$valueCnt) :false; } 解法2:哈希表 设置一个哈希数组,遍历给定数组的元素,如果该元素不在哈希表里,就...
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);//加入映射...
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...
详细的题目描述见上一篇博客《leetcode-137-Single Number II-第一种解法》,这里简单说一下。 有一个数组,所有元素都出现了三次,除了一个元素只出现了一次。输出这个只出现一次的元素。 要求时间复杂度O(n),空间复杂度O(1)。 要完成的函数: int singleNumber(vector<int>& s) ...
Single Number II 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? 首先本能地想到一个算法,可是脑子一转,觉得是要O(n*n)时间复杂度。
LeetCode 136 Single Number(仅仅出现一次的数字) 翻译 给定一个整型数组,除了某个元素外其余元素均出现两次。 1. 找出这个仅仅出现一次的元素。 备注: 你的算法应该是一个线性时间复杂度。 你能够不用额外空间来实现它吗? 原文 Given an array of integers, every element appears twice except for one. Find ...
Leetcode学习(17)—— 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?
leetcode 算法解析(一):260. Single Number III 260.Single Number II 原题链接 本题其实算是比较简单,在 leetcode 上也只是 medium 级别,ac 率也很高,最好先自己尝试,本文只是单纯的记录一下自己整体的思路; 在阅读本文章之前,最好先解锁本题的简单模式136.Single Number,这对理解本题有较大的帮助;...
vector<int>singleNumber(vector<int>& nums){ vector<int> singleV;if(nums.size() <=0)returnsingleV;intresultExclusiveOR =0;for(inti =0; i < nums.size(); ++i) resultExclusiveOR ^= nums[i];unsignedintindexOf1 =FindFirstBigIs1(resultExclusiveOR);intsingleNum1 =0, singleNum2 =0;for...