LeetCode[136] Single Number 位运算异或，HashMap（Java）

Problem：

Thought1：

The standard key to solve this problem is bit manipulation. As XOR will return 1 only when two bits are different, so if two numbers are the same, XOR will return 0. Thus the last remaining number will be the single one.

Java Solution1

public class SingleNumber {
// method 1:Bit manipulation: XOR
public int singleNumber(int[] nums) {
int result = nums[0];
for (int i = 1; i < nums.length; i++) {
result = result ^ nums[i];
}
return result;
}

Thought2：

Since the hint tags show that one of the ways to solve this problem is to utilize the HashTable(HasnMap in Java), so I have also tried this one.

Java Solution2

// method 2: HashMap
public int singleNumberHashMap(int[] nums) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();

for (int i = 0; i < nums.length; i++) {
if (map.containsKey(nums[i])) {
map.remove(nums[i]);
} else {
map.put(nums[i], i);
}
}
// System.out.println(map.entrySet());
for(Map.Entry<Integer, Integer> m : map.entrySet()){
return m.getKey();
}
return 0;
}

Online Judge: Single Number