Leetcode 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:

The order of the result is not important. So in the above example,

[5, 3]

is also correct.

Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

解题思路：

1. 用HashSet ,速度慢。

2. 用Bit Manipulation。

首先计算nums数组中所有数字的异或，记为xor 令lowbit = xor & -xor，lowbit的含义为xor从低位向高位，第一个非0位所对应的数字 例如假设xor = 6（二进制：0110），则-xor为（二进制：1010，-6的补码，two's complement） 则lowbit = 2（二进制：0010）

根据异或运算的性质，“同0异1” 记只出现一次的两个数字分别为a与b，可知a & lowbit与b & lowbit的结果一定不同

通过这种方式，即可将a与b拆分开来。

Java code:

方法一：

public int[] singleNumber(int[] nums) {
Set<Integer> set = new HashSet<Integer>();
int[] result = new int[2];
for(int i = 0; i< nums.length; i++){
if(set.contains(nums[i])){
set.remove(nums[i]);
}else{
set.add(nums[i]);
}
}
Iterator it = set.iterator();
result[0] = (int)it.next();
result[1] = (int)it.next();
return result;
}

方法二：

public int[] singleNumber(int[] nums) {
int record = 0;
for(int num: nums){
record ^= num;
}
record &= -record;
int[] result = {0, 0};
for(int num: nums){
if((num & record) == 0){
result[0] ^= num;
}else{
result[1] ^= num;
}
}
return result;
}

Reference:

1. https://leetcode.com/discuss/52351/c-o-n-time-o-1-space-9-line-solution-with-detail-explanation
2. https://leetcode.com/discuss/52521/share-two-java-solution-%EF%BC%9Ad
3. http://bookshadow.com/weblog/2015/08/17/leetcode-single-number-iii/