leetcode 34. Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return 

[-1, -1]

.

For example,

Given 

[5, 7, 7, 8, 8, 10]

 and target value 8,

return 

[3, 4]

.

int* searchRange(int* nums, int numsSize, int target, int* returnSize) {
if (numsSize == 0 || nums == NULL)
{
*returnSize = 2;
int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
re[0] = -1; re[1] = -1;
return re;
}
if (nums[0] > target || nums[numsSize - 1] < target)
{
*returnSize = 2;
int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
re[0] = -1; re[1] = -1;
return re;
}
if (numsSize == 1)
{
*returnSize = 2;
int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
if (nums[0] == target)
{
re[0] = 0; re[1] = 0; return re;
}
else
{
re[0] = -1; re[1] = -1;
return re;
}
}
int st1 = 0; int end1 = numsSize - 1;
while (end1 > st1)
{
if (nums[end1] > target)
end1 = (st1 + end1) / 2;
else
{
end1 = 2 * end1 - st1 > numsSize - 1 ? numsSize - 1 : 2 * end1 - st1;
st1 = (st1 + end1) / 2;
}
if (nums[end1] == target&&end1 == numsSize - 1)
break;
}
if (nums[end1] != target)
{
*returnSize = 2;
int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
re[0] = -1; re[1] = -1;
return re;
}
int st2 = 0; int end2 = end1;
if (nums[0] == target)
end2 = 0;
else
while (true)
{
if (nums[end2] == target&&nums[st2] != target && (end2 - st2 == 1 || end2 == st2))
break;
if (nums[end2] >= target)
end2 = (st2 + end2) / 2;
else
{
end2 = 2 * end2 - st2 > end1 ? end1 : 2 * end2 - st2;
st2 = (st2 + end2) / 2;
}
}
*returnSize = 2;
int*re = NULL; re = (int*)malloc(sizeof(int)**returnSize);
re[0] = end2; re[1] = end1;
return re;
}

accepted