136. Single Number

use hashmap

T: O(n)

S: O(n)

class Solution {
    public int singleNumber(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0)+1);
        }
        for (int key : map.keySet()) {
            if (map.get(key) == 1) {
                return key;
            }
        }
        return -1;
    }
}j

use bit

一樣的會變 0

T: O(n)

S: O(1)

class Solution {
    public int singleNumber(int[] nums) {
        int res = nums[0];
        for (int i = 1; i < nums.length; i++) {
            res ^= nums[i];
        }
        
        return res;
    }
}

or

class Solution {
    public int singleNumber(int[] nums) {
        int res = 0;
        for (int num : nums) {
            res ^= num;
        }
        return res;
    }
}ja