找到两个数的差为一个特定数

原创于 2017-02-23 00:36:30 发布 · 499 阅读

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探讨如何在已排序数组中找到两个整数a和b，使得它们之间的差值等于给定的整数k。文章提供了一种线性时间复杂度的解决方案，并通过示例代码展示了解决方案的具体实现。

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Given an integer 'k' and an sorted array A (can consist of both +ve/-ve nos), output 2 integers from A such that a-b=k.
PS:
nlogn solution would be to check for the occurence of k-a[i] (using binary search) when you encounter a[i]. methods like hash consume space.

Is an O(n) solution with O(1) extraspace possible?

void fun(vector<int> &a, int key)
{
	int n = a.size();
	if (n < 2)
	{
		return;
	}

	key = abs(key);
	int p = 0;
	int q = 1;
	while (q < n)
	{
		if (a[q] - a[p] == key)
		{
			cout << p << " " << q << endl;
			return;
		}
		else if (a[q] - a[p] > key)
		{
			p++;
		}
		else
		{
			q++;
		}
	}
}