Single Number

SingleNumber II

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?

public class Solution {    public static int singleNumber(int[] nums) {          // if(nums.length == 1) return nums[0];          int result = 0;          for (int i = 0; i < 32; i++) {              int cnt = 0;              int bit = 1 << i;              for (int num : nums) {                  if ((num & bit) != 0) {  等于零不用计 result本来就是零 去掉这个判断的话 cnt每次都等于nums.length没用了                    cnt++;                  }              }              cnt %= 3;              if (cnt == 1) result |= bit;          }          return result;      }  }

public class Solution {    public int singleNumber(int[] nums) {        Arrays.sort(nums);        int i=0;        while(i<nums.length-1){            if(nums[i]==nums[i+2])                i=i+3;             else return nums[i];        }       return nums[i];    }}

SingleNumber I

public class Solution {    public int singleNumber(int[] nums) {        Arrays.sort(nums);        int i=0;        while(i<nums.length-1){            if(nums[i]==nums[i+1])                i=i+2;             else return nums[i];        }       return nums[i];    }}

异或运算具有交换性，1异或2 异或2==1异或（2异或2） 两个相同的数异或得0，任何数与0异或得原值

public class Solution {    public int singleNumber(int[] nums) {       int result=0;       for(int i=0;i<nums.length;i++){           result^=nums[i];       }       return result;    }}

260. Single Number III

 

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QuestionEditorial Solution

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].

public class Solution {    public int[] singleNumber(int[] nums) {        Arrays.sort(nums);        int i=0,j=0;        int [] result=new int[2];        while(i<nums.length-1){            if(nums[i]==nums[i+1])                i=i+2;            else {            result[j++]=nums[i];            i++;            if(j==2) break;            }        }        if(j==1) result[1]=nums[i];        return result;            }}

首先计算nums数组中所有数字的异或，记为xor令lowbit = xor & -xor，lowbit的含义为xor从低位向高位，第一个非0位所对应的数字例如假设xor = 6（二进制：0110），则-xor为（二进制：1010，-6的补码，two's complement）则lowbit = 2（二进制：0010）根据异或运算的性质，“同0异1”记只出现一次的两个数字分别为a与b可知a & lowbit与b & lowbit的结果一定不同通过这种方式，即可将a与b拆分开来

public class Solution {    public int[] singleNumber(int[] nums) {        // Pass 1 :         // Get the XOR of the two numbers we need to find        int diff = 0;        for (int num : nums) {            diff ^= num;        }        // Get its last set bit        diff &= -diff;                // Pass 2 :        int[] rets = {0, 0}; // this array stores the two numbers we will return        for (int num : nums)        {            if ((num & diff) == 0) // the bit is not set            {                rets[0] ^= num;            }            else // the bit is set            {                rets[1] ^= num;            }        }        return rets;    }}