剑指offer-和为s的两个数字VS和为s的连续正数序列

/所给数组为排序数组，若有多组，数组乘积最小的一组
//而距离越远乘积越小，所以从两端开始对数组进行遍历。
bool FindNumbersWithSum(int data[], int length, int sum, int *num1, int *num2)
//参数：给定数组，数组长度，给定和，满足要求的两个数值；
//考虑特殊情况，若不存在和为s的两个数，加入判断的found;
{
bool found = false;
if (length < 1 || num1 == NULL || num2 == NULL)
return found;
int start = 0;
int end = length - 1;
while (end>start)
{
int cur = data[start] + data[end];
if (cur == sum)
{
*num1 = data[start];
*num2 = data[end];
found = true;
}
else if (cur > sum)
{
end--;
}
else
start++;
}
return found;
}

void FindContinuousSequence(int sum)//找到所有和为s的连续序列；
{
if (sum < 3)
return;
int start = 1;
int end = 2;
int middle = (sum + 1) / 2;
int cur = start + end;
while (start < middle)
{
if (cur == sum)
print(start, end);
while (cur >sum&&start < middle)//截止条件判断，cur是否大于sum；
{
cur -= start;
start++;
if (cur == sum)
print(start, end);
}
end++;
cur += end;
}
}
void print(int small, int big)
{
for (int i = small; i <= big; i++)
printf("%d\n", i);
}