Leetcode-136:Single Number

[title3]
[Single Number][/title3]

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

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory

[思路] 1.哈希表：创建一个HashMap，key值为数组元素，value值为出现次数，for循环，若未出现在哈希表中，将此元素put，反之，则remove，最后哈希表中 只剩下出现一次的元素；

2.异或：出现两次的元素都抵消了，只剩下出现一次的元素。

//HashTable
public static int singleNumber(int[] nums) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i =0; i < nums.length; i++) {
if (!map.containsKey(nums[i])) {
map.put(nums[i], 1);

}
else {
map.remove(nums[i]);
}
}
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(nums[i])) {
return nums[i];
}
}
return 0;
}

//异或
public static int singleNumber(int[] nums) {

int res = 0;
for (int i = 0; i < nums.length; i++) {
res = res ^ nums[i];
}
return res;
}