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