【LeetCode】23.Single Number

最新推荐文章于 2024-07-31 16:36:59 发布

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最新推荐文章于 2024-07-31 16:36:59 发布

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博客围绕LeetCode上一道题目展开，题目要求在非空整数数组中找出唯一只出现一次的元素，算法需有线性时间复杂度且不使用额外内存。给出了示例，还进行了算法分析，提到用异或操作去除偶数次数字，最后包含提交代码和测试代码。

题目描述(Easy)

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?

题目链接

https://leetcode.com/problems/single-number/description/

Example 1:

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

Example 2:

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

算法分析

异或操作去除偶数次的数字，时间复杂度 $O(n)$ ，空间复杂度 $O(1)$ 。

提交代码：

class Solution {
public:
	int singleNumber(vector<int>& nums) {
		int result = nums[0];
		for (int i = 1; i < nums.size(); ++i)
			result ^= nums[i];

		return result;
	}
};

测试代码：

// ====================测试代码====================
void Test(const char* testName, vector<int>& nums, int expected)
{
	if (testName != nullptr)
		printf("%s begins: \n", testName);

	Solution s;
	int result = s.singleNumber(nums);

	if (result == expected)
		printf("passed\n");
	else
		printf("failed\n");

}

int main(int argc, char* argv[])
{
	vector<int> ratings = { 2, 2, 1};
	Test("Test1", ratings, 1);

	ratings = { 4, 1, 2, 1, 2 };
	Test("Test2", ratings, 4);
	
	return 0;
}