leetcode 260. Single Number III | 260. 只出现一次的数字 III（位运算：分组异或）

题目

https://leetcode.com/problems/single-number-iii/

题解：分组异或

参考1：讨论区题解
you know you can eliminate doubles with xor so do that first. whats left over is xor of the two single numbers. between those, there must be 1 bit that is set thats unique to that number. set a mask for that bit and xor again with only numbers that have that bit set. you will then find the first number. the first xor is xor of the two numbers. so xor it one more time with the first number to get the other number

参考2：官方题解

class Solution {public int[] singleNumber(int[] nums) {int xor = 0;for (int n : nums)xor ^= n;int a = xor;int b = xor; // int sn = -1; // 方法1：最高位1对应的索引位置 // while (xor != 0) { // xor >>>= 1; // sn++; // } // int mask = 1 << sn; // 选择最高位1所在的位置为maskint mask = 1; // 方法2：选择最低位1所在的位置为maskwhile ((mask & xor) == 0)mask = mask << 1;for (int n : nums) {if ((n & mask) == 0) a ^= n;else b ^= n;}return new int[]{a, b};} }

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