题目：

One Edit Distance

Given two strings S and T, determine if they are both oneedit distance apart.

Hint:

1. If | n – m | is greater than 1, we know immediately both are not one-editdistance apart.

2. It might help if you consider these cases separately, m == n and m ≠ n.

3. Assume that m is always ≤ n, which greatly simplifies the conditionalstatements. If m > n, we could just simply swap S and T.

4. If m == n, it becomes finding if there is exactly one modified operation. Ifm ≠ n, you do not have to consider the delete operation. Just consider theinsert operation in T.

bool OneEditDistance(const string &s1, const string &s2){	if (s1.size() > s2.size())		return OneEditDistance(s2, s1);	if (s2.size() - s1.size()>1)		return false;	int i1 = 0, i2 = 0;	while (i1 < s1.size())	{		if (s1[i1] == s2[i2])		{			++i1;			++i2;		}		else			break;	}	if (i1 == s1.size())		return true;	++i2;	while (i1 < s1.size() && i2