异或应用

1.  给出两个整数a和b, 求他们的和, 但不能使用 + 等数学运算符 （异或又名： 不进位加法）

a + b = (a ^ b) + (a & b << 1)

2. 数组中，只有一个数出现一次，剩下都出现两次，找出出现一次的数

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

3. 数组中，只有一个数出现一次，剩下都出现三次，找出出现一次的数

Ones，Twos俩个数表示当前位的状态

如果Twos里面ith是1，则ith当前为止出现1的次数除以3的状态是2

如果Ones里面ith是1，则ith目前为止出现1的次数除以3的状态是1

outputinputcommentsonesones ^ nums[i]~twos ones =(ones ^ nums[i]) & ~twostwos 0-0-不一样处候选 1110保留 10012+1 = 3 消除twostwos ^ nums[i])~onestwos = (twos ^ nums[i]) & ~onesonescomments 0-0-不一样处候选 11101个数非奇数，是偶数， keep 10011个数奇数，非偶数， 消除

public class Solution {    public int singleNumber(int[] nums) {        int ones = 0, twos = 0;        for(int i = 0; i < nums.length; i++){            ones = (ones ^ nums[i]) & ~twos;            twos = (twos ^ nums[i]) & ~ones;        }        return ones;    }}

4.  数组中，只有两个数出现一次，剩下都出现两次，找出出现一次的这两个数
设此两个数为x和y，对于x和y，一定是其中一个这一位是1，另一个这一位不是1。对于其他成对出现的元素，要么都在x所在的那个集合，要么在y所在的那个集合。

public class Solution {    public int[] singleNumber(int[] nums) {        //用于记录，区分“两个”数组        int diff = 0;        for(int i = 0; i < nums.length; i ++) {            diff ^= nums[i];        }               //diff & (-diff)就是取diff的最后一位1的位置        diff &= -diff;                int[] rets = {0, 0};         for(int i = 0; i < nums.length; i ++) {            //分属两个“不同”的数组            if ((nums[i] & diff) == 0) {                rets[0] ^= nums[i];            }            else {                rets[1] ^= nums[i];            }        }        return rets;    }}

5. Maximum XOR of Two Numbers in an Array

int findMaximumXOR(vector<int>& nums) {        int max = 0, mask = 0;        unordered_set<int> t;        // search from left to right, find out for each bit is there         // two numbers that has different value        for (int i = 31; i >= 0; i--){            // mask contains the bits considered so far            mask |= (1 << i);            t.clear();            // store prefix of all number with right i bits discarded            for (int n: nums){                t.insert(mask & n);            }                        // now find out if there are two prefix with different i-th bit            // if there is, the new max should be current max with one 1 bit at i-th position, which is candidate            // and the two prefix, say A and B, satisfies:            // A ^ B = candidate            // so we also have A ^ candidate = B or B ^ candidate = A            // thus we can use this method to find out if such A and B exists in the set             int candidate = max | (1<<i);            for (int prefix : t){                if (t.find(prefix ^ candidate) != t.end()){                    max = candidate;                    break;                }                            }        }        return max;    }

Sample:

input1010111101100011output1100init0[] 0imasksetcandidate max310001000
'000010001000211001000
1100
0100
001111001100111101010
1110
0110
001011101100011111010
1111
0110
001111011100