Single Number 解答

Question

Given an array of integers, every element appears twice except for one. Find that single one.

Solution 1 -- Set

We can use a hash set to record each integer's appearing time. Time complexity O(n), space cost O(n)

 public class Solution {
     public int singleNumber(int[] nums) {
         Set<Integer> counts = new HashSet<Integer>();
         int length = nums.length, result = nums[0];
         for (int i = 0; i < length; i++) {
             int tmp = nums[i];
             if (counts.contains(tmp))
                 counts.remove(tmp);
             else
                 counts.add(tmp);
         }
         for (int tmp : counts)
             result = tmp;
         return result;
     }
 }

Solution 2 -- Bit Manipulation

The key to solve this problem is bit manipulation. XOR will return 1 only on two different bits. So if two numbers are the same, XOR will return 0. Finally only one number left.

 public int singleNumber(int[] A) {
     int x = 0;
     for (int a : A) {
         x = x ^ a;
     }
     return x;
 }