二分查找的原始寫法如這篇文章所示
如果條件變化一下,比如數組中的元素可能重複,要求返回最小(或最大)的下標。
或者找出數組中第一個大於value(也就是最小的大於value)的元素下標等等,
代碼實現會有所不同。
一共分爲六種不同情況,下面列出這些變形的二分查找
// 找出第一個與value相等的元素
int firstEqual(int A[], int left, int right, int value)
{
int originleft = left, originright = right;
while (left <= right)
{
int mid = left + ((right - left) >> 1);
if (value > A[mid])
left = mid + 1;
else
right = mid - 1;
}
if (A[left] == value && left >= originleft && right <= originright)
return left;
return -1;
}
// 查找第一個大於value的元素
int firstBigger(int A[], int left, int right, int value)
{
while (left <= right)
{
int mid = left + ((right - left) >> 1);
if (value >= A[mid])
left = mid + 1;
else
right = mid - 1;
}
return left;
}
// 查找第一個等於或者大於value的元素
int firstEqualOrBigger(int A[], int left, int right, int value)
{
while (left <= right)
{
int mid = left + ((right - left) >> 1);
if (value > A[mid])
left = mid + 1;
else
right = mid - 1;
}
return left;
}
// 找出最後一個與value相等的元素
int lastEqual(int A[], int left, int right, int value)
{
int originleft = left, originright = right;
while (left <= right)
{
int mid = left + ((right - left) >> 1);
if (value < A[mid])
right = mid - 1;
else
left = mid + 1;
}
if (A[right] == value && left >= originleft && right <= originright)
return right;
return -1;
}
// 查找最後一個小於value的元素
int lastSmaller(int A[], int left, int right, int value)
{
while (left <= right)
{
int mid = left + ((right - left) >> 1);
if (value <= A[mid])
right = mid - 1;
else
left = mid + 1;
}
return right;
}
// 查找最後一個等於或者小於value的元素
int lastEqualOrSmaller(int A[], int left, int right, int value)
{
while (left <= right)
{
int mid = left + ((right - left) >> 1);
if (value < A[mid])
right = mid - 1;
else
left = mid + 1;
}
return right;
}