Contains Duplicate

最新推荐文章于 2022-07-03 13:04:54 发布

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本文介绍两种检查整数数组中是否存在重复元素的方法。第一种方法采用双重遍历对比的方式，时间复杂度为O(n^2)；第二种方法通过哈希表记录每个元素出现次数，时间复杂度降低到O(n)，并详细解释了实现过程。

given an array of integers, find if the array contains any duplicates.
your function should return true if any value appears at least twice in the array,

and it should return false if every element is distinct.

//solution1 time(O(n^2))
bool containsDuplicate1(int* nums, int numsSize) {
	int* p = nums;
	int* q = nums + numsSize - 1;
	int* cursor = NULL;
	bool result = true;
	while (q - p>0)
	{
		cursor = p + 1;
		while (q - cursor >= 0)
		{
			if (*cursor == *p)
				return true;
			else
			{
				++cursor;
			}
		}
		++p;
	}
	return false;

}

//solution2 time(O(n))
bool containsDuplicate2(int* nums, int numsSize) {
	bool result = true;
	if (numsSize == 0)
		result = false;
	int hashkey[numsSize];
	int hashvalue[numsSize];
	for (int k = 0; k != numsSize; ++k)
		hashvalue[k] = 0;

	for (int i = 0; i != numsSize; ++i)
	{
		int id = (nums[i] + numsSize) % numsSize;
		if (hashvalue[id] == 0)
		{
			++hashvalue[id];
			hashkey[id] = nums[i];
		}
		else if (hashkey[id] == nums[i])
			++hashvalue[id];
		else
		{
			while (hashvalue[id] != 0)
			{
				id = (id + 1) % numsSize;
			}
			hashkey[id] = nums[i];
			hashvalue[id] += 1;
		}
	}
	for (int j = 0; j != numsSize; ++j)
	{
		if (hashvalue[j] >= 2)
		{
			result = true;
			break;
		}
		else
			result = false;
	}
	return result;
}