问题描述: 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? 要求:采用线性时间复杂度,并且最好不使用多余的空间. 一.位操作 异或运算,…

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…

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? Subscribe to see which companies asked this…

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. For example: Given nums = [1, 2, 1, 3, 2, 5], return [3, 5]. Note: The order…

Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 这道题是之前那道 Single Number 单独的数字的延伸,那道题的解法…

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? 本来是一道非常简单的题,但是由于加上了时间复杂度必须是O(n),并且空间复杂度为O(1)…

[1]LeetCode 136 Single Number 题意:奇数个数,其中除了一个数只出现一次外,其他数都是成对出现,比如1,2,2,3,3...,求出该单个数. 解法:容易想到异或的性质,两个相同的数异或为0,那么把这串数从头到尾异或起来,最后的数就是要求的那个数. 代码如下: class Solution { public: int singleNumber(vector<int>& nums) { ; ;i<nums.size();i++) sum ^= nums[i…

Given an array of integers, every element appears twice except for one. Find that single one. 思路: 最经典的方法,利用两个相同的数异或结果为0的性质,则将整个数组进行异或,相同的数俩俩异或,最后得到的就是那个single number,复杂度是O(n) 代码: class Solution { public: int singleNumber(vector<int>& nums) { int…

题目地址:here 题目大意:一个整数数组中,只有一个数出现一次,其余数都出现3次,在O(n)时间,O(1)空间内找到这个出现一次的数 对于”只有一个数出现一次,其余数出现2次“的情况,很简单,只要把数组中所有数异或的结果就是这个出现一次的数,leetcode上也有这个题目. 关于本题:我们用一个大小为32的数组cnt来记录:(int数据的每个二进制位1的出现次数)%3,程序中cnt[i]记录第i+1个二进制位.如果某个数出现了三次并且他的第i个二进制位为1,因为做了模3操作,那么该数对数组cn…

Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 没辙,智商碾压题.楼主没遇到之前就只会这种做法. public int si…