Problem:

Given a non-empty 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?

Example 1:

Input: [2,2,1]
Output: 1

Example 2:

Input: [4,1,2,1,2]
Output: 4

Analysis:

本题思路比较简单，使用Map数据结构来记录数组中的每个元素，每当第二次遇见的时候就删除，这样数组最后就剩下一个元素了。代码如下：

 

Code:

class Solution {
    public int singleNumber(int[] nums) {
        Map<Integer, Integer> temp = new HashMap<>();
        for(int ans : nums) {
            if(temp.containsKey(ans)) {
                temp.remove(ans);
            } else {
                temp.put(ans, 1);
            }
        }
        for(Integer ans : temp.keySet()) {
            return ans;
        }
        return 0;
    }
}