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Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
Follow up: Could you implement a solution with a linear runtime complexity and without using extra memory?
Example 1:
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| Input: nums = [2,2,1]
Output: 1
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Example 2:
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| Input: nums = [4,1,2,1,2]
Output: 4
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Example 3:
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| Input: nums = [1]
Output: 1
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Java
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| 位元運算自動轉Binary，XOR is single number fliter
0 xor 0 = 0
0 xor 1 = 1
1 xor 0 = 1
1 xor 1 = 0
解釋
int [] array = {2,2,1}; // result ^= i
byte result;
result = (byte)2^(byte)2; // 0b010^0b010 = 0b000 成對的結果一定是 0
result = (byte)0^(byte)1; // 0b000^0b001 = 0b001
System.out.println(result); // 1
import java.util.*;
class Solution {
public int singleNumber(int[] nums) {
int result = 0;
for(int i : nums) {
result ^= i;
}
return result;
}
}
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