[leetcode]136. Single Number

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

 

 

1.普通解法（利用额外hash）

//hash解法
class Solution {
    public int singleNumber(int[] nums) {
        Map<Integer, Integer> hash = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if(hash.get(nums[i]) == null) {
                hash.put(new Integer(nums[i]), new Integer(0));
            } else {
                hash.put(new Integer(nums[i]), new Integer(1));
            }
        }
        for (int i = 0; i < nums.length; i++) {
            if(hash.get(nums[i]) == 0)
                return nums[i];
        }
        return -1;
    }
}

2.位运算 （按位异或），利用位运算特性

class Solution {
    public int singleNumber(int[] nums) {
        int result = 0;
        for (int i = 0; i < nums.length; i++) {
            result ^= nums[i];
        }
        return result;
        
    }
}