題目

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

題解：分組異或

參考1：討論區(qū)題解
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

參考2：官方題解

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對(duì)應(yīng)的索引位置 // while (xor != 0) { // xor >>>= 1; // sn++; // } // int mask = 1 << sn; // 選擇最高位1所在的位置為maskint mask = 1; // 方法2：選擇最低位1所在的位置為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};} }

總結(jié)

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