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  • 260. Single Number III

    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:

    1. The order of the result is not important. So in the above example, [5, 3] is also correct.
    2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

    Similar

    136. Single Number

    137. Single Number II

    Explain:

    In one's-complement (反码) representation, positive numbers are simply represented as themselves, and negative numbers are represented by the one's complement of their absolute value

    In two's-complement (补码) representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value

    [+1] = [00000001] = [00000001] = [00000001]

    [-1]  = [10000001] = [11111110] = [11111111]

    计算机使用补码进行计算

     1 public class Solution {
     2     public int[] singleNumber(int[] nums) {
     3         // Pass 1 : 
     4         // Get the XOR of the two numbers we need to find
     5         int diff = 0;
     6         for (int num : nums) {
     7             diff ^= num;
     8         }
     9         // Get its last set bit which num1 is diff from num2
    10         diff &= -diff;
    11 
    12         // Pass 2 :
    13         int[] rets = {0, 0}; // this array stores the two numbers we will return
    14         for (int num : nums)
    15         {
    16             if ((num & diff) == 0) // the bit is not set // use ()
    17             {
    18                 rets[0] ^= num;
    19             }
    20             else // the bit is set
    21             {
    22                 rets[1] ^= num;
    23             }
    24         }
    25         return rets;
    26     }
    27 }
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  • 原文地址:https://www.cnblogs.com/joycelee/p/5407799.html
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