136. Single Number

题目

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

Follow up: Could you implement a solution with a linear runtime complexity and without using extra memory?

Example 1:

Input: nums = [2,2,1] Output: 1 Example 2:

Input: nums = [4,1,2,1,2] Output: 4 Example 3:

Input: nums = [1] Output: 1

Constraints:

1 <= nums.length <= 3 * 104 -3 * 104 <= nums[i] <= 3 * 104 Each element in the array appears twice except for one element which appears only once.

题解

位运算

异或运算有以下几个特点：

一个数和 0 做 XOR 运算等于本身：a⊕0 = a

一个数和其本身做 XOR 运算等于 0：a⊕a = 0

XOR 运算满足交换律和结合律：a⊕b⊕a = (a⊕a)⊕b = 0⊕b = b

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

哈希表

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class Solution {
    public int singleNumber(int[] nums) {
        HashMap<Integer, Integer> hs = new HashMap<>();
        for (int num : nums) {
            hs.put(num, hs.getOrDefault(num, 0) + 1);
        }
        int single = 0;
        for (Map.Entry<Integer, Integer> entry : hs.entrySet()) {
            if (entry.getValue() == 1) {
                single = entry.getKey();
            }
        }
        return single;
    }
}